Maharashtra Board Class 12 Physics Chapter 1 Mechanical Properties of Fluids Exercise Solutions

1. Choose the correct option.

(i) When seen from below, the blades of a ceiling fan are seen to be revolving anticlockwise and their speed is decreasing. Select the correct statement about the directions of its angular velocity and angular acceleration.
(a) Angular velocity upwards, angular acceleration downwards.
(b) Angular velocity downwards, angular acceleration upwards.
(c) Both, angular velocity and angular acceleration, upwards.
(d) Both, angular velocity and angular acceleration, downwards.
Answer: (a) Angular velocity upwards, angular acceleration downwards.
In simple words: Imagine the fan is spinning and you use your right hand to follow the motion; your thumb points up (velocity). Since it is slowing down, the 'braking' force or acceleration must point in the opposite direction, which is down.

📝 Teacher's Note: Use the Right-Hand Thumb Rule to demonstrate direction. Remind students that if an object is slowing down, the acceleration vector is always opposite to the velocity vector.

🎯 Exam Tip: Pay close attention to the perspective (seen from below vs. above) as it changes the perceived direction of rotation.

(ii) A particle of mass 1 kg, tied to a 1.2 m long string is whirled to perform the vertical circular motion, under gravity. The minimum speed of a particle is 5 m/s. Consider the following statements.
P) Maximum speed must be \( 5\sqrt{5} \) m/s.
Q) Difference between maximum and minimum tensions along the string is 60 N.
Select the correct option.
(a) Only statement P is correct.
(b) Only statement Q is correct.
(c) Both the statements are correct.
(d) Both the statements are incorrect.
Answer: (c) Both the statements are correct.
In simple words: In a vertical loop, the object moves fastest at the bottom and slowest at the top. The pull on the string (tension) is also much stronger at the bottom than at the top.

📝 Teacher's Note: Explain the conservation of energy in vertical circular motion. For statement Q, show that the tension difference \( T_{max} - T_{min} = 6mg \).

🎯 Exam Tip: Remember the standard formula for tension difference in a vertical circle is always \( 6mg \) regardless of the velocity, provided the motion is completed.

(iii) Select the correct statement about the formula (expression) of the moment of inertia (M.I.) in terms of mass M of the object and some of its distance parameter/s, such as R, L, etc.
(a) Different objects must have different expressions for their M.I.
(b) When rotating about their central axis, a hollow right circular cone and a disc have the same expression for the M.I.
(c) Expression for the M.I. for a parallelepiped rotating about the transverse axis passing through its centre includes its depth.
(d) Expression for M.I. of a road and that of a plane sheet is the same about a transverse axis.
Answer: (b) When rotating about their central axis, a hollow right circular cone and a disc have the same expression for the M.I.
In simple words: Moment of inertia depends on how mass is distributed. Even though a cone and a disc look different, the way their mass is spread out relative to the middle axis can be described by the same math formula.

📝 Teacher's Note: This is a great time to show students a table of M.I. for various shapes. Highlight that both a uniform disc and a hollow cone (of a specific type) share the formula \( I = \frac{1}{2}MR^2 \).

🎯 Exam Tip: Don't assume unique shapes have unique formulas. Always refer to the standard derivation for the axis specified.

(iv) In a certain unit, the radius of gyration of a uniform disc about its central and transverse axis is \( \sqrt{2.5} \). Its radius of gyration about a tangent in its plane (in the same unit) must be
(a) \( \sqrt{5} \)
(b) 2.5
(c) \( 2\sqrt{2.5} \)
(d) \( \sqrt{12.5} \)
Answer: (b) 2.5
In simple words: Radius of gyration is a special distance that helps us calculate how hard it is to spin something. Moving the spinning point to the edge (tangent) changes this distance according to specific geometry rules.

📝 Teacher's Note: Guide students through the Parallel Axis Theorem. First find the radius of gyration about the diameter, then apply the theorem to find it about the tangent.

🎯 Exam Tip: Be careful to identify which axis the "radius of gyration" refers to (central vs. diameter vs. tangent) before starting your calculation.

(v) Consider following cases:
(P) A planet revolving in an elliptical orbit.
(Q) A planet revolving in a circular orbit.
Principle of conservation of angular momentum comes in force in which of these?
(a) Only for (P)
(b) Only for (Q)
(c) For both, (P) and (Q)
(d) Neither for (P), nor for (Q)
Answer: (c) For both, (P) and (Q)
In simple words: Since gravity pulls directly toward the center, there is no 'twisting' force (torque) acting on the planet. This means its spinning momentum stays constant whether the path is a circle or an oval.

📝 Teacher's Note: Relate this to Kepler's Second Law (Areal Velocity). Explain that gravitational force is a central force, and the torque of a central force about the center is always zero.

🎯 Exam Tip: Angular momentum is conserved in any "central force" field. This is a very common conceptual question.

Vi) A thin walled hollow cylinder is rolling down an incline, without slipping. At any instant, the ratio ”Rotational K.E.: Translational K.E.: Total K.E.” is
(a) 1:1:2
(b) 1:2:3
(c) 1:1:1
(d) 2:1:3
Answer: (a) 1:1:2
In simple words: For a thin pipe rolling down a hill, the energy is perfectly split: half goes into moving the pipe forward and the other half goes into making it spin.

📝 Teacher's Note: Use the formula for kinetic energy of a rolling body: \( E = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \). For a thin hollow cylinder, \( I = MR^2 \), making the two terms equal.

🎯 Exam Tip: Remember the ratios for common shapes: Ring/Hollow Cylinder (1:1:2), Disc/Solid Cylinder (1:2:3), Solid Sphere (2:5:7).

2. Answer in brief.

(i) Why are curved roads banked?
Answer: A car while taking a turn performs circular motion. If the road is level (or horizontal road), the necessary centripetal force is the force of static friction between the car tyres and the road surface. The friction depends upon the nature of the surfaces in contact and the presence of oil and water on the road. If the friction is inadequate, a speeding car may skid off the road. Since the friction changes with circumstances, it cannot be relied upon to provide the necessary centripetal force. Moreover, friction results in fast wear and tear of the tyres.
To avoid the risk of skidding as well as to reduce the wear and tear of the car tyres, the road surface at a bend is tilted inward, i.e., the outer side of the road is raised above its inner side. This is called banking of road. On a banked road, the resultant of the normal reaction and the gravitational force can act as the necessary centripetal force. Thus, every car can be safely driven on such a banked curve at certain optimum speed, without depending on friction. Hence, a road should be properly banked at a bend.
The angle of banking is the angle of inclination of the surface of a banked road at a bend with the horizontal.
In simple words: Roads are tilted at turns so the road itself helps push the car toward the inside of the curve. This prevents sliding when it's slippery and keeps the tires from wearing out too fast.

📝 Teacher's Note: Use a diagram to show how the normal force (\( N \)) splits into vertical (\( N \cos \theta \)) and horizontal (\( N \sin \theta \)) components. The horizontal component provides the centripetal force.

🎯 Exam Tip: Mentioning "unreliability of friction" and "wear and tear of tires" are key points that examiners look for to award full marks.

(ii) Do we need a banked road for a two wheeler? Explain.
Answer: When a two-wheeler takes a turn along an unbanked road, the force of friction provides the centripetal force. The two-wheeler leans inward to counteract a torque that tends to topple it outward. Firstly, friction cannot be relied upon to provide the necessary centripetal force on all road conditions. Secondly, the friction results in wear and tear of the tyres. On a banked road at a turn, any vehicle can negotiate the turn without depending on friction and without straining the tyres.
In simple words: While bike riders can lean to turn, a banked road is still better because it provides a natural push toward the center, making the turn safer and saving the tires from damage.

📝 Teacher's Note: Explain the difference between leaning (rider's action) and banking (road's design). Leaning changes the center of gravity relative to the contact point to prevent toppling.

🎯 Exam Tip: Contrast the "leaning" of a bike with the "banking" of a road to show you understand both concepts.

(iii) On what factors does the frequency of a conical pendulum depend? Is it independent of some factors?
Answer: The frequency of a conical pendulum, of string length L and semivertical angle \( \theta \), is
\( n = \frac{1}{2\pi} \sqrt{\frac{g}{L \cos \theta}} \)
where g is the acceleration due to gravity at the place.
From the above expression, we can see that
1. \( n \propto \sqrt{g} \)
2. \( n \propto \frac{1}{\sqrt{L}} \)
3. \( n \propto \frac{1}{\sqrt{\cos \theta}} \) (if \( \theta \) increases, \( \cos \theta \) decreases and \( n \) increases)
4. The frequency is independent of the mass of the bob.
In simple words: How fast the pendulum swings in a circle depends on gravity, how long the string is, and the angle of the swing. It does not matter if the weight at the bottom is heavy or light.

📝 Teacher's Note: Emphasize the "independence of mass." Students often assume a heavier bob will swing differently, which is a common misconception.

🎯 Exam Tip: Always state the formula first when asked about "factors," then list the proportionalities clearly.

(iv) Why is it useful to define radius of gyration?
Answer: Definition: The radius of gyration of a body rotating about an axis is defined as the distance between the axis of rotation and the point at which the entire mass of the body can be supposed to be concentrated so as to give the same moment of inertia as that of the body about the given axis.
The moment of inertia (MI) of a body about a given rotation axis depends upon:
1. the mass of the body and
2. the distribution of mass about the axis of rotation.
These two factors can be separated by expressing the MI as the product of the mass (M) and the square of a particular distance (k) from the axis of rotation. This distance is called the radius of gyration and is defined as given above. Thus,
\( I = \sum m_i r_i^2 = Mk^2 \)

\( \implies k = \sqrt{I/M} \)
Physical significance: The radius of gyration is less if I is less, i.e., if the mass is distributed close to the axis; and it is more if I is more, i.e., if the mass is distributed away from the axis. Thus, it gives the idea about the distribution of mass about the axis of rotation.
In simple words: It's like finding a single average distance for all the mass in a spinning object. It tells us at a glance if most of the object's weight is near the center or far out on the edges.

📝 Teacher's Note: Compare "radius of gyration" for a ring vs. a disc to illustrate the concept of mass distribution. A ring has a larger \( k \) because its mass is further from the axis.

🎯 Exam Tip: Define \( k \) mathematically (\( I=Mk^2 \)) and describe its physical significance (distribution of mass) to get full marks.

(v) A uniform disc and a hollow right circular cone have the same formula for their M.I., when rotating about their central axes. Why is it so?
Answer: The radius of gyration of a thin ring of radius \( R_r \) about its transverse symmetry axis is
\( k_r = \sqrt{I_{CM}/M_r} = \sqrt{R_r^2} = R_r \)
The radius of gyration of a thin disc of radius \( R_d \) about its transverse symmetry axis is
\( k_d = \sqrt{I_{CM}/M_d} = \sqrt{\frac{M_d R_d^2/2}{M_d}} = \frac{1}{\sqrt{2}} R_d \)
Given \( k_r = k_d \),
\( R_r = \frac{1}{\sqrt{2}} R_d \) or, equivalently, \( R_d = \sqrt{2} R_r \).
In simple words: Even though they look like different shapes, the math shows that the mass of a disc and the mass of a hollow cone are spread out away from the center in exactly the same way.

📝 Teacher's Note: Point out that the distribution of mass relative to the central axis is what matters, not the 3D height of the object. A hollow cone can be thought of as a stack of rings with varying radii.

🎯 Exam Tip: When explaining "why," focus on the mass distribution being mathematically identical for the chosen axis of rotation.

Question 3. While driving along an unbanked circular road, a two-wheeler rider has to lean with the vertical. Why is it so? With what angle the rider has to lean? Derive the relevant expression. Why such a leaning is not necessary for a four wheeler?
Answer: When a bicyclist takes a turn along an unbanked road, the force of friction \( f_s \) provides the centripetal force; the normal reaction of the road \( \vec{N} \) is vertically up. If the bicyclist does not lean inward, there will be an unbalanced outward torque about the centre of gravity, \( f_s \cdot h \), due to the friction force that will topple the bicyclist outward. The bicyclist must lean inward to counteract this torque (and not to generate a centripetal force) such that the opposite inward torque of the couple formed by \( \vec{N} \) and the weight \( \vec{g} \), \( mg \cdot a = f_s \cdot h_1 \).
Since the force of friction provides the centripetal force,
\( f_s = \frac{mv^2}{r} \)
If the cyclist leans from the vertical by an angle \( \theta \), the angle between \( \vec{N} \) and \( \vec{F} \):
\( \tan \theta = \frac{f_s}{N} = \frac{mv^2/r}{mg} = \frac{v^2}{gr} \)
Hence, the cyclist must lean by an angle
\( \theta = \tan^{-1} \left( \frac{v^2}{gr} \right) \)
When a car takes a turn along a level road, apart from the risk of skidding off outward, it also has a tendency to roll outward due to an outward torque about the centre of gravity due to the friction force. But a car is an extended object with four wheels. So, when the inner wheels just get lifted above the ground, it can be counterbalanced by a restoring torque of the couple formed by the normal reaction (on the outer wheels) and the weight.
In simple words: A biker leans so gravity can balance out the 'push' that tries to flip them outward during a turn. Cars don't need to do this because they are wider and have four wheels to stay stable.

📝 Teacher's Note: Explain that for a bicycle, there is only one line of contact with the ground. For a car, the width between the wheels provides a base that can create a "restoring torque" without the driver needing to tilt the whole vehicle.

🎯 Exam Tip: The derivation of \( \tan \theta = v^2/gr \) is essential. Make sure to clearly state that the leaning is to "counteract torque."

Question 4. Using the energy conservation, derive the expressions for the minimum speeds at different locations along a vertical circular motion controlled by gravity. Is zero speed possible at the uppermost point? Under what condition/s? Also prove that the difference between the extreme tensions (or normal forces) depends only upon the weight of the object.
Answer: In a non uniform vertical circular motion, e.g., those of a small body attached to a string or the loop-the-loop manoeuvers of an aircraft or motorcycle or skateboard, the body must have some minimum speed to reach the top and complete the circle. In this case, the motion is controlled only by gravity and zero speed at the top is not possible. However, in a controlled vertical circular motion, e.g., those of a small body attached to a rod or the giant wheel (Ferris wheel) ride, the body or the passenger seat can have zero speed at the top, i.e., the motion can be brought to a stop.
[Derivation: Using conservation of energy between top and bottom points]
\( \frac{1}{2}mv_{top}^2 + mg(2r) = \frac{1}{2}mv_{bottom}^2 \)
For string: \( v_{top} = \sqrt{gr} \), \( v_{bottom} = \sqrt{5gr} \), \( v_{mid} = \sqrt{3gr} \)
Difference in tensions: \( T_{bottom} - T_{top} = 6mg \)
In simple words: To stay in a circle, you must go fast enough so the string doesn't go slack at the top. The difference in pull on the string between the top and bottom is always exactly 6 times the object's weight.

📝 Teacher's Note: Contrast the "string" case with the "rod" case. For a rod, the speed at the top can be zero because the rod can push up to support the weight, unlike a string which would go limp.

🎯 Exam Tip: Be sure to distinguish between motion controlled by "gravity" (string) and "controlled" motion (rod) when answering about zero speed at the top.

Question 5. Discuss the necessity of radius of gyration. Define it. On what factors does it depend and it does not depend? Can you locate some similarity between the centre of mass and radius of gyration? What can you infer if a uniform ring and a uniform disc have the same radius of gyration?
Answer: Definition: The radius of gyration of a body rotating about an axis is defined as the distance between the axis of rotation and the point at which the entire mass of the body can be supposed to be concentrated so as to give the same moment of inertia as that of the body about the given axis.
The moment of inertia (MI) of a body about a given rotation axis depends upon:
1. the mass of the body and
2. the distribution of mass about the axis of rotation. These two factors can be separated by expressing the MI as the product of the mass (M) and the square of a particular distance (k) from the axis of rotation. This distance is called the radius of gyration.
\( I = \sum m_i r_i^2 = Mk^2 \)

\( \implies k = \sqrt{I/M} \)
Physical significance: It gives the idea about the distribution of mass about the axis of rotation.
Similarity with CM: The centre of mass (CM) locates a point where the entire mass M can be thought to be concentrated for translational motion (\( \vec{F}_{net} = M\vec{a}_{CM} \)). Similarly, radius of gyration locates a point where mass can be concentrated for rotational motion (\( \vec{\tau}_{net} = I\vec{\alpha} \)).
Inference for ring and disc: If a uniform ring (radius \( R_r \)) and disc (radius \( R_d \)) have same radius of gyration \( k \), then \( R_d = \sqrt{2} R_r \). This means the disc must have a larger radius than the ring to have the same "spinning resistance."
In simple words: Radius of gyration is like a single number that summarizes how spread out an object is. Just as the "Center of Mass" is a shortcut for position, the "Radius of Gyration" is a shortcut for spinning.

📝 Teacher's Note: The comparison between CM and \( k \) is a powerful way to help students understand that rotation is just the "angular version" of linear motion concepts.

🎯 Exam Tip: When comparing ring and disc, show the mathematical step \( R_r = R_d/\sqrt{2} \) to demonstrate a complete understanding.

Question 6. State the conditions under which the theorems of parallel axes and perpendicular axes are applicable. State the respective mathematical expressions.
Answer: The theorem of parallel axis is applicable to any body of arbitrary shape. The moment of inertia (MI) of the body about an axis through the centre mass should be known, say, \( I_{CM} \). Then, the theorem can be used to find the MI, I, of the body about an axis parallel to the above axis. If the distance between the two axes is h,
\( I = I_{CM} + Mh^2 \) ...(1)
The theorem of perpendicular axes is applicable to a plane lamina only. The moment of inertia \( I_z \) of a plane lamina about an axis-the z axis- perpendicular to its plane is equal to the sum of its moments of inertia \( I_x \) and \( I_y \) about two mutually perpendicular axes x and y in its plane and through the point of intersection of the perpendicular axis and the lamina.
\( I_z = I_x + I_y \) .... (2)
In simple words: The Parallel Axis theorem works for any 3D shape, but the Perpendicular Axis theorem only works for flat objects like a sheet of paper. They help us calculate how hard it is to spin something around different points.

📝 Teacher's Note: Stress that the perpendicular axis theorem *cannot* be used for 3D objects like spheres or cylinders. This is a very common student mistake.

🎯 Exam Tip: Clearly state "plane lamina only" for the perpendicular axes theorem to avoid losing marks on the applicability condition.

Question 7. Derive an expression that relates angular momentum with the angular velocity of a rigid body.
Answer: Consider a rigid body rotating with a constant angular velocity \( \vec{\omega} \) about an axis through the point O. All the particles of the body perform uniform circular motion about the axis of rotation with the same angular velocity \( \vec{\omega} \). Suppose that the body consists of N particles of masses \( m_1, m_2, ..., m_n \), situated at perpendicular distances \( r_1, r_2, ..., r_n \), respectively from the axis of rotation.
The particle of mass \( m_1 \) revolves along a circle of radius \( r_1 \), with a linear velocity of magnitude \( v_1 = r_1\omega \). The magnitude of the linear momentum of the particle is
\( p_1 = m_1 v_1 = m_1 r_1 \omega \)
The angular momentum of the particle about the axis of rotation is by definition,
\( \vec{L}_1 = \vec{r}_1 \times \vec{p}_1 \)
\( \implies L_1 = r_1 p_1 \sin \theta \)
In this case, \( \theta = 90^\circ \), \( \sin \theta = 1 \)
\( \implies L_1 = r_1 p_1 = r_1 m_1 r_1 \omega = m_1 r_1^2 \omega \)
Similarly \( L_2 = m_2 r_2^2 \omega \), \( L_3 = m_3 r_3^2 \omega \), etc.
The total angular momentum \( L = L_1 + L_2 + ... + L_N \)
\( \implies L = (m_1 r_1^2 + m_2 r_2^2 + ... + m_N r_N^2) \omega \)
\( \implies L = I\omega \)
In vector form, \( \vec{L} = I\vec{\omega} \).
In simple words: This derivation shows that total spinning momentum is simply the object's mass distribution (Moment of Inertia) multiplied by how fast it's spinning.

📝 Teacher's Note: Draw a simple blob shape with an axis through it to represent the rigid body. Label a few points \( m_1, m_2, m_3 \) with distances \( r_1, r_2, r_3 \) to help students visualize the sum.

🎯 Exam Tip: Mention that all particles have the same angular velocity \( \omega \) but different linear velocities \( v \). This is the key logic step.

Question 8. Obtain an expression relating the torque with angular acceleration for a rigid body.
Answer: A torque acting on a body produces angular acceleration. Consider a rigid body rotating about an axis passing through the point O. Suppose that a torque \( \vec{\tau} \) on the body produces uniform angular acceleration \( \vec{\alpha} \) along the axis of rotation.
The body is made up of N particles with masses \( m_1, m_2, ..., m_N \) at distances \( r_1, r_2, ..., r_N \).
The torque \( \tau_1 \) on particle \( m_1 \) is \( \vec{\tau}_1 = \vec{r}_1 \times \vec{F}_1 \)
\( \implies \tau_1 = r_1 F_1 \sin 90^\circ = r_1 F_1 \)
Now, \( F_1 = m_1 a_1 = m_1 r_1 \alpha \)
\( \implies \tau_1 = r_1 (m_1 r_1 \alpha) = m_1 r_1^2 \alpha \)
Similarly, \( \tau_2 = m_2 r_2^2 \alpha \), ..., \( \tau_N = m_N r_N^2 \alpha \)
Total torque \( \tau = \tau_1 + \tau_2 + ... + \tau_N \)
\( \implies \tau = (m_1 r_1^2 + m_2 r_2^2 + ... + m_N r_N^2) \alpha \)
\( \implies \tau = I\alpha \)
In vector form, \( \vec{\tau} = I\vec{\alpha} \).
In simple words: Just as Force = Mass × Acceleration for straight-line motion, Torque = Moment of Inertia × Angular Acceleration for spinning motion.

📝 Teacher's Note: This is the rotational equivalent of Newton's Second Law. Use this analogy to help students remember the formula.

🎯 Exam Tip: State that \( \alpha \) is constant for all particles in a rigid body. This is a crucial assumption in the derivation.

Question 9. State and explain the principle of conservation of angular momentum. Use a suitable illustration. Do we use it in our daily life? When?
Answer: Law (or principle) of conservation of angular momentum: The angular momentum of a body is conserved if the resultant external torque on the body is zero.
Explanation: Since \( L = I\omega \), if \( \tau = 0 \), then \( I_1\omega_1 = I_2\omega_2 \). This means if the moment of inertia decreases, the angular velocity must increase.
(1) Ice dance: A figure skater starts a spin with arms and one leg extended (large \( I \)). When she pulls her limbs close to her body, her moment of inertia decreases significantly, causing her spinning speed (\( \omega \)) to increase greatly to conserve angular momentum.
(2) Diving: A diver tucked into a compact shape has a small \( I \) and spins fast. When entering the water, they stretch out to increase \( I \), which slows down the rotation for a safe entry.
In simple words: If nothing is twisting you from the outside, you will spin faster if you pull your arms in and slower if you stretch them out.

📝 Teacher's Note: Use the "spinning chair" demonstration if possible. Have a student hold weights out and then pull them in to feel the sudden increase in speed.

🎯 Exam Tip: Always specify the condition: "if the resultant external torque is zero." Without this phrase, the definition is incomplete.

Question 10. Discuss the interlink between translational, rotational and total kinetic energies of a rigid object that rolls without slipping.
Answer: Consider a symmetric rigid body rolling on a surface. Its motion can be treated as translation of the CM and rotation about an axis through the CM.
Total Kinetic Energy \( E = E_{tran} + E_{rot} \) ...(1)
Let M and R be the mass and radius. \( \omega \), k and I are angular speed, radius of gyration and MI.
\( v = \omega R \) and \( I = Mk^2 \)
\( E_{tran} = \frac{1}{2}Mv^2 \) and \( E_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}Mk^2(\frac{v^2}{R^2}) \)
\( E = \frac{1}{2}Mv^2 + \frac{1}{2}Mk^2 \frac{v^2}{R^2} \)
\( \implies E = \frac{1}{2}Mv^2 \left( 1 + \frac{k^2}{R^2} \right) \)
This shows total energy is the translational energy multiplied by a factor depending on the shape (\( k^2/R^2 \)).
In simple words: When a ball rolls, some energy makes it move forward and some makes it spin. The total energy is the sum of both.

📝 Teacher's Note: Help students realize that "rolling" is a combination of two motions happening at once. The ratio \( k^2/R^2 \) is the key to identifying how the energy is split.

🎯 Exam Tip: Be comfortable with the conversion \( \omega = v/R \). This is only valid for "rolling without slipping."

Question 11. A rigid object is rolling down an inclined plane. Derive expressions for the acceleration along the track and the speed after falling through a certain vertical distance.
Answer: Consider a body rolling down an incline of height h and angle \( \theta \). By conservation of energy:
\( PE_{initial} = KE_{final} \)
\( Mgh = \frac{1}{2}Mv^2 \left( 1 + \frac{k^2}{R^2} \right) \)

\( \implies v^2 = \frac{2gh}{1 + k^2/R^2} \)

\( \implies v = \sqrt{\frac{2gh}{1 + k^2/R^2}} \)
For acceleration 'a' along length L: \( v^2 = 2aL \)
\( \frac{2gh}{1 + k^2/R^2} = 2a \left( \frac{h}{\sin \theta} \right) \)

\( \implies a = \frac{g \sin \theta}{1 + k^2/R^2} \)
In simple words: When something rolls down a hill, gravity's energy gets divided between moving and spinning. This makes rolling objects speed up more slowly than objects that just slide.

📝 Teacher's Note: Ask students: "Which reaches the bottom first, a solid sphere or a hollow one?" This derivation provides the answer (\( k^2/R^2 \) is smaller for the solid sphere, so its 'a' and 'v' are higher).

🎯 Exam Tip: Practice deriving both 'v' and 'a' as they are frequently asked together in long-form questions.

Question 12. Somehow, an ant is stuck to the rim of a bicycle wheel of diameter 1 m. While the bicycle is on a central stand, the wheel is set into rotation and it attains the frequency of 2 rev/s in 10 seconds, with uniform angular acceleration. Calculate
(i) Number of revolutions completed by the ant in these 10 seconds.
(ii) Time taken by it for first complete revolution and the last complete revolution.
[Ans: 10 rev., \( t_{first} = \sqrt{10} \) s, \( t_{last} = 0.5132 \) s]
Answer: Data: \( r = 0.5 \) m, \( \omega_0 = 0 \), \( \omega = 2 \) rps, \( t = 10 \) s
(i) Average angular speed \( \omega_{av} = \frac{\omega_0 + \omega}{2} = \frac{0 + 2}{2} = 1 \) rps
Total revolutions \( \theta = \omega_{av} \cdot t = 1 \times 10 = 10 \) revolutions.
(ii) Angular acceleration \( \alpha = \frac{\omega - \omega_0}{t} = \frac{2-0}{10} = 0.2 = \frac{1}{5} \) rev/s²
Using \( \theta = \frac{1}{2} \alpha t^2 \):
For 1st rev (\( \theta = 1 \)): \( 1 = \frac{1}{2} \left( \frac{1}{5} \right) t_1^2 \implies t_1^2 = 10 \implies t_1 = \sqrt{10} = 3.162 \) s.
For 9 revs: \( 9 = \frac{1}{2} \left( \frac{1}{5} \right) t_9^2 \implies t_9^2 = 90 \implies t_9 = \sqrt{90} = 9.486 \) s.
Time for 10th rev = \( t_{10} - t_9 = 10 - 9.486 = 0.514 \) s.
In simple words: The ant speeds up along with the wheel. We use the average speed to find total turns, and the acceleration formula to find how time intervals get shorter as it goes faster.

📝 Teacher's Note: Remind students that they can work in "revolutions" and "rps" directly instead of converting to radians, which often simplifies the math in these types of problems.

🎯 Exam Tip: For the "last revolution" part, calculate the time for \( (N) \) revolutions and subtract the time for \( (N-1) \) revolutions.

Question 13. Coefficient of static friction between a coin and a gramophone disc is 0.5. Radius of the disc is 8 cm. Initially the centre of the coin is 2 cm away from the centre of the disc. At what minimum frequency will it start slipping from there? By what factor will the answer change if the coin is almost at the rim?
(use \( g = \pi^2 \) m/s²)
[Ans: 2.5 rev/s, \( n_2 = \frac{1}{2} n_1 \)]
Answer: Data: \( \mu_s = 0.5 \), \( r_1 = 2 \) cm \( = 2 \times 10^{-2} \) m, \( r_2 = 8 \) cm \( = 8 \times 10^{-2} \) m, \( g = \pi^2 \) m/s²
Condition for no slipping: \( mr\omega^2 \leq \mu_s mg \implies \omega^2 \leq \frac{\mu_s g}{r} \)
\( 4\pi^2 f^2 = \frac{\mu_s g}{r} \)
\( f_{min,1}^2 = \frac{\mu_s g}{4\pi^2 r_1} = \frac{0.5 \times \pi^2}{4\pi^2 \times 0.02} = \frac{0.5}{0.08} = 6.25 \)
\( \implies f_{min,1} = \sqrt{6.25} = 2.5 \) rps.
Since \( f \propto 1/\sqrt{r} \):
\( \frac{f_2}{f_1} = \sqrt{\frac{r_1}{r_2}} = \sqrt{\frac{2}{8}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \)
The frequency decreases by a factor of 2 (i.e., it slips at a lower speed when further out).
In simple words: Friction has to hold the coin in place. The further out the coin is, the harder it is for friction to keep it on the spinning disc.

📝 Teacher's Note: This problem demonstrates why items fly off a merry-go-round more easily when you are on the outside edge. The required centripetal force increases with radius for a constant angular speed.

🎯 Exam Tip: Using the proportionality \( f \propto 1/\sqrt{r} \) is a much faster way to solve the second part of the question than recalculating from scratch.

Question 14. Part of a racing track is to be designed for a radius of curvature of 72 m. We are not recommending the vehicles to drive faster than 216 kmph. With what angle should the road be tilted? At what height will its outer edge be, with respect to the inner edge if the track is 10 m wide?
[Ans: \( \theta = \tan^{-1}(5) = 78.69^\circ \), \( h = 9.8 \) m]
Answer: Data: \( r = 72 \) m, \( v = 216 \) km/h \( = 216 \times \frac{5}{18} = 60 \) m/s, \( w = 10 \) m, \( g = 10 \) m/s²
Banking angle: \( \tan \theta = \frac{v^2}{rg} = \frac{60^2}{72 \times 10} = \frac{3600}{720} = 5 \)
\( \implies \theta = \tan^{-1}(5) \approx 78^\circ 4' \)
This is the required angle of banking.
sin θ = h/w
∴ h = w sin θ = (10) sin 78°4′ = 10 × 0.9805
= 9.805 m
This gives the height of the outer edge of the track relative to the inner edge
In simple words: To safely handle cars going at very high speeds, the track needs to be tilted very steeply. For a 10m wide track, the outer edge must be nearly 10m higher than the inner edge!

📝 Teacher's Note: Mention that \( 78^\circ \) is an extremely steep angle for a road, usually only seen on specialized high-speed testing tracks or velodromes.

🎯 Exam Tip: Always convert km/h to m/s before plugging into the banking formula. Failing to do this is the most common reason for wrong answers.

Question 15. The road in the example 14 above is constructed as per the requirements. The coefficient of static friction between the tyres of a vehicle on this road is 0.8, will there be any lower speed limit? By how much can the upper speed limit exceed in this case?
[Ans: \( v_{min} \cong 88 \text{ kmph} \), no upper limit as the road is banked for \( \theta > 45^{\circ} \)]
Answer: Data : \( r = 72 \text{ m} \), \( \theta = 78^{\circ}4' \), \( \mu_s = 0.8 \), \( g = 10 \text{ m/s}^2 \), \( \tan \theta = \tan 78^{\circ}4' = 5 \)
\( v_{min} = \sqrt{rg \left( \frac{\tan \theta - \mu_s}{1 + \mu_s \tan \theta} \right)} \)
\( \implies \) \( v_{min} = \sqrt{(72)(10) \left( \frac{5 - 0.8}{1 + (0.8)(5)} \right)} \)
\( \implies \) \( v_{min} = \sqrt{720 \times \frac{4.2}{5}} = \sqrt{144 \times 4.2} = 12 \times 2.049 \)
\( \implies \) \( v_{min} = 24.588 \text{ m/s} = 88.52 \text{ km/h} \)
This will be the lower limit or minimum speed on this track.
Since the track is heavily banked, \( \theta > 45^{\circ} \), there is no upper limit or maximum speed on this track.
In simple words: When a road is tilted very steeply (more than 45 degrees), a car needs to maintain a minimum speed just to avoid sliding down into the inner edge. However, because the tilt is so high, there is no physical limit to how fast it can go without sliding up, as long as the tires can handle it.

📝 Teacher's Note: Use a steep funnel or a bowl analogy. If you spin a marble slowly, it falls to the bottom; if you spin it fast, it stays up. At very high angles, the "upward" friction actually helps keep the car on track.

🎯 Exam Tip: Note the condition \( \tan \theta > 1 \) (or \( \theta > 45^{\circ} \)). In such cases, if \( \mu_s \tan \theta \ge 1 \), the denominator for the upper speed limit formula becomes zero or negative, indicating no theoretical maximum speed.

Question 16. During a stunt, a cyclist (considered to be a particle) is undertaking horizontal circles inside a cylindrical well of radius 6.05 m. If the necessary friction coefficient is 0.5, how much minimum speed should the stunt artist maintain? Mass of the artist is 50 kg. If she/he increases the speed by 20%, how much will the force of friction be?
[Ans: \( v_{min} = 11 \text{ m/s} \), \( f_s = mg = 500 \text{ N} \)]
Answer: Data : \( r = 6.05 \text{ m} \), \( \mu_s = 0.5 \), \( g = 10 \text{ m/s}^2 \), \( m = 50 \text{ kg} \), \( \Delta v = 20\% \)
\( v_{min} = \sqrt{\frac{rg}{\mu_s}} = \sqrt{\frac{6.05 \times 10}{0.5}} = \sqrt{12.1 \times 10} = \sqrt{121} = 11 \text{ m/s} \)
This is the required minimum speed. So long as the cyclist is not sliding, at every instant, the force of static friction is \( f_s = mg = (50)(10) = 500 \text{ N} \).
In simple words: In the "Well of Death," the wall pushes against the bike to provide turning force, and friction holds the bike up against gravity. As long as the bike goes fast enough not to fall, the friction exactly balances the weight, regardless of how much faster the rider goes.

📝 Teacher's Note: Clarify that while centripetal force depends on speed (\( mv^2/r \)), the vertical friction force only needs to balance the weight (\( mg \)) to prevent falling. Increasing speed increases the normal reaction, not the required friction.

🎯 Exam Tip: Be careful! Many students think friction increases with speed here. Remember that friction only balances the constant weight (\( mg \)) as long as the minimum speed condition is met.

Question 17. A pendulum consisting of a massless string of length 20 cm and a tiny bob of mass 100 g is set up as a conical pendulum. Its bob now performs 75 rpm. Calculate kinetic energy and increase in the gravitational potential energy of the bob. (Use \( \pi^2 = 10 \))
[Ans: \( \cos \theta = 0.8 \), \( K.E. = 0.45 \text{ J} \), \( \Delta(P.E.) = 0.04 \text{ J} \)]
Answer: Data : \( L = 0.2 \text{ m} \), \( m = 0.1 \text{ kg} \), \( n = \frac{75}{60} = \frac{5}{4} \text{ rps} \), \( g = 10 \text{ m/s}^2 \), \( \pi^2 = 10 \),
\( T = \frac{1}{n} = \frac{4}{5} \text{ s} = 0.8 \text{ s} \)
\( T = 2\pi \sqrt{\frac{L \cos \theta}{g}} \)
\( \implies \) \( T^2 = 4\pi^2 \frac{L \cos \theta}{g} \)
\( \implies \) \( h = L \cos \theta = \frac{g T^2}{4\pi^2} = \frac{(10)(0.8)^2}{4(10)} = 0.16 \text{ m} \) ... (1)
\( \implies \) \( \cos \theta = \frac{0.16}{0.2} = 0.8 \)
\( \implies \) \( \theta = \cos^{-1} 0.8 = 36.87^{\circ} = 36^{\circ}5' \)
\( v^2 = rg \tan \theta = (L \sin \theta) (g) \tan 36.87^{\circ} \)
\( \implies \) \( v^2 = (0.12) (10) (0.7500) = 0.9 \)
The \( KE \) of the bob \( = \frac{1}{2} m v^2 = \frac{1}{2} (0.1) (0.9) = 0.045 \text{ J} \)
The increase in gravitational \( PE \),
\( \Delta PE = mg(L - h) = (0.1) (10) (0.2 - 0.16) = 0.04 \text{ J} \)
In simple words: When the pendulum swings in a circle, the bob rises slightly. We calculate how high it rises using its speed and then find the energy it gained from that height and movement.

📝 Teacher's Note: Show the derivation of \( h = L \cos \theta \) on the board. Explain that as the speed (frequency) increases, the bob swings "wider" and "higher," decreasing \( h \).

🎯 Exam Tip: Ensure you convert units to SI (cm to m, g to kg, rpm to rps) before starting calculations to avoid magnitude errors.

Question 18. A motorcyclist (as a particle) is undergoing vertical circles inside a sphere of death. The speed of the motorcycle varies between 6 m/s and 10 m/s. Calculate diameter of the sphere of death. What are the minimum values are possible for these two speeds?
[Ans: \( \text{Diameter} = 3.2 \text{ m} \), \( (v_1)_{min} = 4 \text{ m/s} \), \( (v_2)_{min} = 4\sqrt{5} \text{ m/s} \)]
Answer: Data : \( v_{top} = 6 \text{ m/s} \), \( v_{bot} = 10 \text{ m/s} \), \( g = 10 \text{ m/s}^2 \)
\( v_{bot}^2 = v_{top}^2 + 4gr \)
\( \implies \) \( r = \frac{v_{bot}^2 - v_{top}^2}{4g} = \frac{(10)^2 - (6)^2}{4 \times 10} = \frac{64}{40} = 1.6 \text{ m} \)
The diameter of the sphere of death \( = 3.2 \text{ m} \).
For this \( r \), \( v_{min} = \sqrt{gr} \) at the top.
\( \implies \) \( v_{min} = \sqrt{10 \times 1.6} = \sqrt{16} = 4 \text{ m/s} \)
The corresponding minimum speed at the bottom
\( \implies \) \( \sqrt{5gr} = \sqrt{5(10)(1.6)} = \sqrt{80} = 4\sqrt{5} \text{ m/s} \)
The required minimum values of the speeds are \( 4 \text{ m/s} \) and \( 4\sqrt{5} \text{ m/s} \).
In simple words: To keep a bike moving in a vertical loop, gravity pulls it down at the top, so it needs a certain speed to not fall. We use the difference in speed between the top and bottom to figure out how big the loop is.

📝 Teacher's Note: Emphasize the conservation of energy principle: \( \frac{1}{2}mv_{bot}^2 = \frac{1}{2}mv_{top}^2 + mgh \), where \( h = 2r \).

🎯 Exam Tip: Remember the standard critical speeds for vertical circles: \( \sqrt{gr} \) at the top and \( \sqrt{5gr} \) at the bottom.

Question 19. A metallic ring of mass 1 kg has moment of inertia 1 kg m² when rotating about one of its diameters. It is molten and remoulded into a thin uniform disc of the same radius. How much will its moment of inertia be, when rotated about its own axis.
[Ans: \( 1 \text{ kg m}^2 \)]
Answer: The \( MI \) of the thin ring about its diameter,
\( I_{ring} = \frac{1}{2} MR^2 = 1 \text{ kg.m}^2 \)
Since the ring is melted and recast into a thin disc of same radius \( R \), the mass of the disc equals the mass of the ring \( = M \).
The \( MI \) of the thin disc about its own axis (i.e., transverse symmetry axis) is
\( I_{disc} = \frac{1}{2} MR^2 = I_{ring} \)
\( \implies \) \( I_{disc} = 1 \text{ kg.m}^2 \)
In simple words: A ring's inertia when spinning like a flipped coin is the same formula as a flat disc's inertia when spinning like a wheel. Since mass and radius didn't change, the "effort" to spin them stays the same.

📝 Teacher's Note: This is a conceptual check on formulas. Remind students that for a ring, \( I_{axis} = MR^2 \), but \( I_{diameter} = \frac{1}{2}MR^2 \). For a disc, \( I_{axis} = \frac{1}{2}MR^2 \).

🎯 Exam Tip: Always identify the specific axis of rotation mentioned in the question, as the formula for Moment of Inertia changes with the axis.

Question 20. A big dumb-bell is prepared by using a uniform rod of mass 60 g and length 20 cm. Two identical solid thermocol spheres of mass 25 g and radius 10 cm each are at the two ends of the rod. Calculate moment of inertia of the dumbbell when rotated about an axis passing through its centre and perpendicular to the length.
[Ans: \( 24000 \text{ g cm}^{-2} \)]
Answer: Data : \( M_{sph} = 50 \text{ g} \), \( R_{sph} = 10 \text{ cm} \), \( M_{rod} = 60 \text{ g} \), \( L_{rod} = 20 \text{ cm} \)
The \( MI \) of a solid sphere about its diameter is
\( I_{sph,CM} = \frac{2}{5} M_{sph} R_{sph}^2 \)
The distance of the rotation axis (transverse symmetry axis of the dumbbell) from the centre of sphere, \( h = 30 \text{ cm} \).
The \( MI \) of a solid sphere about the rotation axis, \( I_{sph} = I_{sph, CM} + M_{sph} h^2 \)
For the rod, the rotation axis is its transverse symmetry axis through \( CM \).
The \( MI \) of a rod about this axis,
\( I_{rod} = \frac{1}{12} M_{rod} L_{rod}^2 \)
Since there are two solid spheres, the \( MI \) of the dumbbell about the rotation axis is
\( I = 2 I_{sph} + I_{rod} \)
\( \implies \) \( I = 2 M_{sph} \left( \frac{2}{5} R_{sph}^2 + h^2 \right) + \frac{1}{12} M_{rod} L_{rod}^2 \)
\( \implies \) \( I = 2(50) \left[ \frac{2}{5}(10)^2 + (30)^2 \right] + \frac{1}{12}(60)(20)^2 \)
\( \implies \) \( I = 100(40 + 900) + 5(400) = 94000 + 2000 = 96000 \text{ g}\cdot\text{cm}^2 \)
In simple words: To find the total resistance to spinning, we add the rod's spinning effort and the two spheres' effort. Because the spheres are far from the center, we use a special rule (Parallel Axis Theorem) to calculate their contribution.

📝 Teacher's Note: Explain the Parallel Axis Theorem clearly. The distance \( h \) is the distance from the center of the rod to the center of the sphere (\( L/2 + R \)).

🎯 Exam Tip: Watch out for mass values. The question says "25 g" but the provided solution uses "50 g". Always double-check if mass is given for one sphere or the total. In exams, use the value provided in the question text.

Question 21. A flywheel used to prepare earthenware pots is set into rotation at 100 rpm. It is in the form of a disc of mass 10 kg and radius 0.4 m. A lump of clay (to be taken equivalent to a particle) of mass 1.6 kg falls on it and adheres to it at a certain distance x from the centre. Calculate x if the wheel now rotates at 80 rpm.
[Ans: \( x = \frac{1}{\sqrt{8}} \text{ m} = 0.35 \text{ m} \)]
Answer: Data : \( f_1 = 60 \text{ rpm} = 60/60 \text{ rot/s} = 1 \text{ rot/s} \),
\( f_2 = 30 \text{ rpm} = 30/60 \text{ rot/s} = \frac{1}{2} \text{ rot/s} \), \( \Delta E = -100 \text{ J} \)
(i) Rotational \( KE \), \( E = \frac{1}{2} I \omega^2 = \frac{1}{2} I (2\pi f)^2 = 2\pi^2 I f^2 \)
The change in \( KE \), \( \Delta E = E_2 - E_1 = 2\pi^2 I (f_2^2 - f_1^2) \)
\( \implies \) \( I = \frac{\Delta E}{2\pi^2(f_2^2 - f_1^2)} = \frac{-100}{2(3.142)^2 [(\frac{1}{2})^2 - 1^2]} = \frac{-100}{2(3.142)^2 (-\frac{3}{4})} = \frac{200}{3(3.142)^2} = 6.753 \text{ kg.m}^2 \)
This gives the \( MI \) of the flywheel about the given axis.

(ii) Angular momentum, \( L = I \omega = I(2\pi f) = 2\pi I f \)
The change in angular momentum, \( \Delta L = L_2 - L_1 = 2\pi I (f_2 - f_1) \)
\( \implies \) \( \Delta L = 2 \times 3.142 \times 6.753 (\frac{1}{2} - 1) = -3.142 \times 6.753 = -21.22 \text{ kg.m}^2/\text{s} \)
In simple words: When clay drops on a spinning wheel, the wheel slows down because it now has more mass to carry. We use the change in speed to find where exactly the clay landed.

📝 Teacher's Note: The provided answer text in the source contains different data values (\( 60 \text{ rpm}, 30 \text{ rpm}, -100 \text{ J} \)) compared to the question (\( 100 \text{ rpm}, 80 \text{ rpm} \)). Use this as a teaching moment to show how to adapt formulas to given data.

🎯 Exam Tip: In problems involving sticking mass, apply the Principle of Conservation of Angular Momentum (\( I_1 \omega_1 = I_2 \omega_2 \)).

Question 22. Starting from rest, an object rolls down along an incline that rises by 3 units in every 5 units (along it). The object gains a speed of \( \sqrt{10} \text{ m/s} \) as it travels a distance of \( \frac{5}{3} \text{ m} \) along the incline. What can be the possible shape/s of the object?
[Ans: \( \frac{K^2}{R^2} = 1 \). Thus, a ring or a hollow cylinder]
Answer: Data : \( \sin \theta = \frac{3}{5} \), \( u = 0 \), \( v = \sqrt{10} \text{ m/s} \), \( L = \frac{5}{3} \text{ m} \), \( g = 10 \text{ m/s}^2 \)
\( v = \sqrt{\frac{2gL \sin \theta}{1 + (K^2/R^2)}} = \sqrt{\frac{2gL \sin \theta}{1 + \beta}} \)
\( \implies \) \( v^2 = \frac{2gL \sin \theta}{1 + \beta} \)
\( \implies \) \( 1 + \beta = \frac{2gL \sin \theta}{v^2} = \frac{2(10)(\frac{5}{3})(\frac{3}{5})}{(\sqrt{10})^2} = \frac{20}{10} = 2 \)
\( \implies \) \( \beta = \frac{K^2}{R^2} = 1 \)
Therefore, the body rolling down is either a ring or a cylindrical shell.
In simple words: Different shapes roll down hills at different speeds because some use more energy to spin than others. By measuring the speed, we calculated a "shape number" (\( \beta \)). A value of 1 belongs to rings and hollow tubes.

📝 Teacher's Note: Explain that the ratio \( K^2/R^2 \) tells us how the mass is distributed. For a ring it's 1, for a solid disc it's 0.5, and for a sphere it's 0.4.

🎯 Exam Tip: Memorize the \( K^2/R^2 \) values for common objects: Ring=1, Disc=1/2, Solid Sphere=2/5, Hollow Sphere=2/3.

Question. Attach a body of suitable mass to a spring balance so that it stretches by about half its capacity. Now whirl the spring balance so that the body performs a horizontal circular motion. You will notice that the balance now reads more for the same body. Can you explain this ?
Answer: Due to outward centrifugal force.
In simple words: When you spin something in a circle, it wants to fly outward. This outward "pull" (centrifugal force) adds to the pull of gravity, making the spring stretch more than when the object is just hanging still.

📝 Teacher's Note: This is a great demonstration for "pseudo-forces." In the frame of the spring, the centrifugal force acts radially outward, increasing the tension.

🎯 Exam Tip: Always mention "centrifugal force" when explaining observations from a rotating frame of reference.

Question. Obtain the condition for not toppling (rollover) for a four-wheeler. On what factors does it depend and how?
Answer: Consider a car of mass \( m \) taking a turn of radius \( r \) along a level road. As seen from an inertial frame of reference, the forces acting on the car are :
1. the lateral limiting force of static friction \( \vec{f_s} \) on the wheels-acting along the axis of the wheels and towards the centre of the circular path- which provides the necessary centripetal force,
2. the weight \( m\vec{g} \) acting vertically downwards at the centre of gravity (C.G.)
3. the normal reaction \( \vec{N} \) of the road on the wheels, acting vertically upwards effectively at the C.G. Since maximum centripetal force = limiting force of static friction,
\( ma_r = \frac{mv^2}{r} = f_s \) .... (1)
In a simplified rigid-body vehicle model, we consider only two parameters—the height \( h \) of the C.G. above the ground and the average distance \( b \) between the left and right wheels called the track width.
The friction force \( f_s \) on the wheels produces a torque \( \tau_t \) that tends to overturn/rollover the car about the outer wheel. Rotation about the front-to-back axis is called roll.
\( \tau_t = f_s \cdot h = \left( \frac{mv^2}{r} \right) h \) ... (2)
When the inner wheel just gets lifted above the ground, the normal reaction \( \vec{N} \) of the road acts on the outer wheels but the weight continues to act at the C.G. Then, the couple formed by the normal reaction and the weight produces a opposite torque \( \tau_r \) which tends to restore the car back on all four wheels
\( \tau_r = mg \cdot \frac{b}{2} \) .... (3)
The car does not topple as long as the restoring torque \( \tau_r \) counterbalances the toppling torque \( \tau_t \). Thus, to avoid the risk of rollover, the maximum speed that the car can have is given by
\( \left( \frac{mv^2}{r} \right) h = mg \cdot \frac{b}{2} \implies v_{max} = \sqrt{\frac{rbg}{2h}} \) ... (4)
Thus, vehicle tends to roll when the radial acceleration reaches a point where inner wheels of the four-wheeler are lifted off of the ground and the vehicle is rotated outward. A rollover occurs when the gravitational force passes through the pivot point of the outer wheels, i.e., the C.G. is above the line of contact of the outer wheels. Equation (3) shows that this maximum speed is high for a car with larger track width and lower centre of gravity.
In simple words: A car topples if it turns so fast that the "sideways push" from friction overcomes the car's stability. Stability depends on how wide the car is (track width) and how low it sits (center of gravity). Wider, lower cars are much harder to flip.

📝 Teacher's Note: Use a cereal box on a tray to demonstrate. Tilt the tray or slide it fast to show how height (\( h \)) and width (\( b \)) affect when it tips over.

🎯 Exam Tip: In the derivation, clearly state that for "just toppling," the normal reaction from the inner wheels becomes zero. The final formula \( v_{max} = \sqrt{\frac{rbg}{2h}} \) is crucial for numericals.

Question. Think about the normal reactions. Where are those and how much are those?
Answer: In a simplified vehicle model, we assume the normal reactions to act equally on all the four wheels, i.e., \( mg/4 \) on each wheel. However, the C.G. is not at the geometric centre of a vehicle and the wheelbase (i.e., the distance \( L \) between its front and rear wheels) affects the weight distribution of the vehicle. When a vehicle is not accelerating, the normal reactions on each pair of front and rear wheels are, respectively,
\( N_f = \frac{d_r}{L} mg \) and \( N_r = \frac{d_f}{L} mg \)
where \( d_r \) and \( d_f \) are the distances of the rear and front axles from the C.G.
In simple words: Gravity doesn't always push down on all tires equally. If the heavy engine is in the front, the front tires have a stronger "normal reaction" (push back from the road) than the back tires.

📝 Teacher's Note: Relate this to a see-saw. If the pivot (C.G.) is closer to one end, that end supports more of the weight. This is why front-engine cars have different tire wear than mid-engine cars.

🎯 Exam Tip: Note that during acceleration or braking, these reactions change, causing the car to "pitch" forward or backward. Mention this "dynamic loading" for extra marks.

Question. What is the recommendations on loading a vehicle for not toppling easily?
Answer: Overloading (or improper load distribution) or any load placed on the roof raises a vehicle’s centre of gravity, and increases the vehicle’s likelihood of rolling over. A roof rack should be fitted by considering weight limits.
Road accidents involving rollovers show that vehicles with higher \( h \) (such as SUVs, pickup vans and trucks) topple more easily than cars. Untripped rollovers normally occur when a top-heavy vehicle attempts to perform a panic manoeuver that it physically cannot handle.
In simple words: To keep a car stable, keep the heavy stuff low. Putting heavy luggage on a roof rack makes the car "top-heavy," which makes it much easier to flip over during a sharp turn.

📝 Teacher's Note: Discuss the safety of SUVs vs. Sedans. SUVs have higher clearance (higher \( h \)), making them statistically more prone to rollovers if not driven carefully.

🎯 Exam Tip: Use terms like "lowering the Centre of Gravity" and "optimizing the static stability factor" to show technical understanding.

Question. If a vehicle topples while turning, which wheels leave the contact with the road? Why?
Answer: Inner wheels.
Friction provides the centripetal force for turning, but it also creates a torque about the outer wheels that tends to lift the inner wheels. As speed increases, this lifting torque eventually overcomes the weight's restoring torque.
In simple words: When you turn sharply to the left, the car wants to lean to the right. If it goes too fast, the left (inner) wheels will lift off the ground first because the whole car is trying to "pivot" over the outside tires.

📝 Teacher's Note: This is an application of moments. The pivot point during toppling is the line of contact of the outer wheels.

🎯 Exam Tip: Clearly identify the pivot point (outer wheels) to explain why the inner wheels are the ones that lift.

Question. How does [tendency to] toppling affect the tyres?
Answer: While turning, shear stress acts on the tyre-road contact area. Due to this, the treads and side wall of a tyre deform. Apart from less control, this contributes to increased and uneven wear of the shoulder of the tyres.
Each wheel is placed under a small inward angle (called camber) in the vertical plane. Under severe lateral acceleration, when the car rolls, the camber angle ensures the complete contact area is in contact with the road and the wheels are now in vertical position. This improves the cornering behavior of the car. Improperly inflated and worn tyres can be especially dangerous because they inhibit the ability to maintain vehicle control.
In simple words: Sharp turns literally try to peel the rubber sideways. This wears out the edges (shoulders) of the tires faster. Engineers tilt the wheels slightly (camber) so that when the car leans during a turn, the tire ends up perfectly flat against the road for the best grip.

📝 Teacher's Note: Show images of "Camber angle." Negative camber is often seen in racing cars to maximize grip during high-speed cornering.

🎯 Exam Tip: Mention "shear stress" and "uneven shoulder wear" as the primary physical impacts on tires during cornering.

Question. What is the recommendation for this?
Answer: Because of uneven wear of the tyre shoulders, tyres should be rotated every 10000 km-12000 km. To avoid skidding, rollover and tyre-wear, the force of friction should not be relied upon to provide the necessary centripetal force during cornering. Instead, the road surface at a bend should be banked, i.e., tilted inward.
In simple words: Move your tires around (rotation) so they wear out evenly. More importantly, roads should be tilted at corners so the road itself helps the car turn, reducing the "scrubbing" effect on the tires.

📝 Teacher's Note: Explain that banking allows the normal reaction to provide the centripetal force, reducing the "burden" on friction and thus saving the tires.

🎯 Exam Tip: "Banking of roads" is the engineering solution to minimize tire wear and maximize safety at curves.

Question. Determine the angle to be made with the vertical by a two-wheeler while turning on a horizontal track?
Answer: When a bicyclist takes a turn along an unbanked road, the force of friction \( \vec{f_s} \) provides the centripetal force; the normal reaction of the road \( \vec{N} \) is vertically up. If the bicyclist does not lean inward, there will be an unbalanced outward torque about the centre of gravity, \( f_s \cdot h \), due to the friction force that will topple the bicyclist outward. The bicyclist must lean inward to counteract this torque such that the opposite inward torque of the couple formed by \( \vec{N} \) and the weight \( \vec{g} \), \( mg \cdot a = f_s \cdot h_1 \)
Since the force of friction provides the centripetal force, \( f_s = \frac{mv^2}{r} \)
If the cyclist leans from the vertical by an angle \( \theta \),
\( \tan \theta = \frac{f_s}{N} = \frac{mv^2/r}{mg} = \frac{v^2}{gr} \)
Hence, the cyclist must lean by an angle \( \theta = \tan^{-1} \left( \frac{v^2}{gr} \right) \).
In simple words: A bike rider leans into a turn to stay balanced. The lean creates a "pull" that stops the bike from flipping over outward. The faster the turn, the more the rider has to lean.

📝 Teacher's Note: Contrast this with a car. A car cannot "lean" its body significantly, which is why it needs four wheels for stability or a banked road. A two-wheeler uses the rider's shift in CG.

🎯 Exam Tip: The angle \( \theta \) is with the *vertical*. Sometimes questions ask for the angle with the horizontal, which would be \( 90 - \theta \).

Question. We have mentioned about ‘static friction’ between road and tyres. Why is it static friction? What about kinetic friction between road and tyres?
Answer: When a car takes a turn on a level road, the point of contact of the wheel with the surface is instantaneously stationary if there is no slipping. Hence, the lateral force on the car is the limiting force of static friction between the tyres and road. As long as the wheels are rolling, there is lateral force of static friction and longitudinal force of rolling friction.
If the car skids, the friction force is kinetic friction; more importantly, the direction of the friction force then changes abruptly from lateral to that opposite the velocity of skidding and not towards the centre of the curve, so that the car cannot continue in its curved path.
In simple words: As a tire rolls, the part touching the ground for a split second is actually standing still relative to the road. This is why we use "static" friction. "Kinetic" friction only happens when you start sliding or skidding, and at that point, you lose control of the turn.

📝 Teacher's Note: Use a rolling coin to show that the bottom point isn't sliding against the table. Kinetic friction is much less predictable and usually lower than static friction, leading to loss of control during skids.

🎯 Exam Tip: Emphasize that "rolling without slipping" involves static friction. Kinetic friction is associated with skidding/sliding.

Question. What do you do if your vehicle is trapped on a slippery or sandy road? What is the physics involved?
Answer: Avoid excessive spinning of your tyres because it will most likely just dig you into a deeper hole. Momentum is the key to getting unstuck. One method is the rocking method—rocking your car backwards and forwards to gain momentum. Your best option is usually to gain traction and momentum by wedging a car mat (or sticks, leaves, gravel or rocks) in front and under your drive wheels. Once you start moving, keep the momentum going until you are on more solid terrain.
In simple words: When stuck in sand, spinning the tires fast just drills a hole because there's no grip (friction). By putting a mat or branches under the tires, you create a rough surface that the tires can "grab" onto using friction to move the car forward.

📝 Teacher's Note: This is a practical application of increasing the coefficient of friction (\( \mu \)) by changing the contact surface. Rocking uses the concept of impulse and momentum.

🎯 Exam Tip: Mention that adding external materials like gravel or mats increases the "coefficient of static friction," which provides the necessary horizontal force to move.

Use your brain power (Textbook Page No. 6)

Question 1. As a civil engineer, you are to construct a curved road in a ghat. In order to calculate the banking angle \( \theta \), you need to decide the speed limit. How will you decide the values of speed and radius of curvature at the bend ?
Answer: For Indian roads, Indian Road Congress (IRC) specifies the design speed depending on the classification of roads (national highways, district roads, etc.) and terrain. For the radius of curvature at a bend, IRC specifies the absolute minimum values based on the minimum design speed.
A civil engineer refers to banking as superelevation \( e \); \( e = \tan \theta \). The sequence of design usually goes like this :
1. Knowing the design speed \( V \) and radius \( r \), calculate the superelevation for 75% of design speed ignoring friction : \( e = \frac{V^2}{225r} \)
2. If \( e < 0.07 \), consider this calculated value. If \( e > 0.07 \), then take \( e = 0.07 \).
3. Check the value of \( \mu \) for \( e = 0.07 \) at full design speed. If \( \mu < 0.15 \), then \( e = 0.07 \) is safe.
In simple words: Engineers don't just guess speed; they follow national safety rules. They design the tilt of the road so that at 75% of the expected speed, a car could turn safely even if the road was made of ice (zero friction).

📝 Teacher's Note: This explains why some turns feel "natural" at certain speeds. The 75% rule is a safety buffer for various road conditions and driver behaviors.

🎯 Exam Tip: Mention "Superelevation" and the standard 75% design speed rule used by civil engineers for banking calculations.

Use your brain power (Textbook Page No. 7)

Question 1. If friction is zero, can a vehicle move on the road? Why are we not considering the friction in deriving the expression for the banking angle?
Answer: Friction is necessary for any form of locomotion. Without friction, a vehicle cannot move. The banking angle for a road at a bend is calculated for optimum speed at which every vehicle can negotiate the bend without depending on friction to provide the necessary lateral centripetal force.
In simple words: You need friction just to start or stop a car (like walking on ice). However, we design road tilt assuming zero friction as a safety measure so that even on a rainy or oily day, a car can turn safely at the recommended speed.

📝 Teacher's Note: Differentiate between "starting/stopping" (longitudinal friction) and "turning" (lateral friction). Banking specifically replaces the need for lateral friction.

🎯 Exam Tip: Use the term "Optimum Speed" (\( v = \sqrt{rg \tan \theta} \)) to describe the speed where no friction is required for turning.

Use your brain power (Textbook Page No. 12)

Question 1. What is expected to happen if one travels fast over a speed breaker? Why?
Answer: The maximum speed with which a car can travel over a road surface, which is in the form of a convex arc of radius \( r \), is \( \sqrt{rg} \) where \( g \) is the acceleration due to gravity. For a speed breaker, \( r \) is very small (of the order of 1 m). Hence, one must slow down considerably while going over a speed breaker. Otherwise, the car will lose contact with the road and land with a thud.
In simple words: A speed breaker is like a tiny hill. If you go too fast, your car's momentum will launch it into the air because gravity isn't strong enough to pull it down fast enough to follow the curve of the bump.

📝 Teacher's Note: This is a great example of the Normal Reaction (\( N \)) becoming zero. When \( mv^2/r = mg \), the tires no longer press against the ground.

🎯 Exam Tip: The condition for "losing contact" is \( N = 0 \). Use the formula \( v \le \sqrt{rg} \) to explain safety limits on convex surfaces.

Question 2. How does the normal force on a concave suspension bridge change when a vehicle is travelling on it with a constant speed ?
Answer: At the lowest point, \( N - mg \) provides the centripetal force. Therefore, \( N - mg = \frac{mv^2}{r} \), so that \( N = m(g + \frac{v^2}{r}) \).
Therefore, \( N \) increases with increasing \( v \).
In simple words: When you bottom out on a sagging bridge, you feel "heavier." This is because the road has to push up extra hard not just to hold your weight, but also to change your direction from going down to going up.

📝 Teacher's Note: Compare this to a convex bridge where you feel "lighter" (\( N = m(g - v^2/r) \)). This explains why suspension systems are stressed most at the bottom of a dip.

🎯 Exam Tip: For concave surfaces (dips), the Normal force is *always* greater than the weight (\( N > mg \)).

Use your brain power (Textbook Page No. 15)

Question 1. For the point P in above, we had to extend OC to Q to meet the perpendicular PQ. What will happen to the expression for I if the point P lies on OC?
Answer: There will be no change in the expression for the \( MI \) (I) about the parallel axis through O.
In simple words: The Parallel Axis Theorem works regardless of where the specific point \( P \) is located within the object; the total sum of all points will always result in the same final formula.

📝 Teacher's Note: This refers to the derivation of \( I = I_0 + Mh^2 \). The geometric arrangement of a single point \( P \) doesn't change the general validity of the theorem for the entire mass distribution.

🎯 Exam Tip: The Parallel Axis Theorem is independent of the shape of the body, as long as one axis passes through the Center of Mass.