JEE Physics System Of Particles and Rotational Motion MCQs Set 67

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Multiple Choice Questions

Question 1: The torque \( \vec\tau \) on a body about a given point is found to be \( \vec A\times\vec L \), where \( \vec A \) is a constant vector and \( \vec L \) is the angular momentum of the body about that point. Which of the following statements are correct? (a) \( d\vec L/dt \) is perpendicular to \( \vec L \) at all instants of time. (b) The component of \( \vec L \) in the direction of \( \vec A \) does not change with time. (c) The magnitude of \( \vec L \) does not change with time. (d) \( \vec L \) does not change with time.
(a) (a) and (d) only
(b) (b) and (d) only
(c) (a), (b), (c) and (d)
(d) (a), (b) and (c) only
Show Answer & Explanation

Answer: (d) (a), (b) and (c) only

Explanation:
1. \( \dfrac{d\vec L}{dt}=\vec A\times\vec L \) is perpendicular to both \( \vec A \) and \( \vec L \): \( (a) \) true.
2. \( \dfrac{d}{dt}(\vec L\cdot\vec A)=\vec A\cdot(\vec A\times\vec L)=0 \): \( (b) \) true. \( \dfrac{d}{dt}L^2=2\vec L\cdot(\vec A\times\vec L)=0 \): \( (c) \) true.
3. \( d\vec L/dt\neq0 \) in general (the direction of \( L \) changes), so \( (d) \) is false.

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Question 2: A smooth slot is cut along a diameter of a horizontal disc of radius R rotating counter-clockwise with constant angular speed \( \omega \) about its vertical axis through the centre. A small block of mass m is gently placed in the slot at distance R/2 from the centre at t = 0 and is constrained to move only along the slot. The distance r of the block from the centre at time t is
(a) \( \dfrac R4\left(e^{\omega t}+e^{-\omega t}\right) \)
(b) \( \dfrac R2\cos2\omega t \)
(c) \( \dfrac R2\cos\omega t \)
(d) \( \dfrac R4\left(e^{2\omega t}+e^{-2\omega t}\right) \)
Show Answer & Explanation

Answer: (a) \( \dfrac R4\left(e^{\omega t}+e^{-\omega t}\right) \)

Explanation:
1. Along the slot the only radial acceleration is centrifugal: \( \ddot r=\omega^2r \).
2. General solution \( r=Ae^{\omega t}+Be^{-\omega t} \) with \( r(0)=R/2 \) and \( \dot r(0)=0 \) gives \( A=B=R/4 \).
3. \( r=\dfrac R4\left(e^{\omega t}+e^{-\omega t}\right) \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Question 3: For the block in the smooth slot of the rotating disc described above (slot along the x-axis in the disc frame, disc rotating about the z-axis with \( \vec\omega=\omega\hat k \)), the net reaction of the disc on the block is
(a) \( m\omega^2R\sin\omega t\,\hat j-mg\hat k \)
(b) \( \tfrac12m\omega^2R\left(e^{\omega t}-e^{-\omega t}\right)\hat j+mg\hat k \)
(c) \( \tfrac12m\omega^2R\left(e^{2\omega t}-e^{-2\omega t}\right)\hat j+mg\hat k \)
(d) \( -m\omega^2R\cos\omega t\,\hat j-mg\hat k \)
Show Answer & Explanation

Answer: (b) \( \tfrac12m\omega^2R\left(e^{\omega t}-e^{-\omega t}\right)\hat j+mg\hat k \)

Explanation:
1. The slot is smooth, so the horizontal reaction is perpendicular to the slot (along \( y \)) and supplies the Coriolis term \( 2m\,v_{rot}\,\omega \).
2. \( v_{rot}=\dot r=\dfrac{R\omega}4\left(e^{\omega t}-e^{-\omega t}\right) \), so \( N_y=2m\omega\dot r=\tfrac12m\omega^2R\left(e^{\omega t}-e^{-\omega t}\right) \).
3. Vertically the reaction balances weight: \( +mg\hat k \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Question 4: A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to rotate about its axis. Two massless spring toy-guns, each carrying a steel ball of mass 0.05 kg, are attached to the platform at a distance 0.25 m from the centre on its either side along a diameter. Each gun simultaneously fires its ball horizontally and perpendicular to the diameter in opposite directions. After leaving the platform, the balls have a horizontal speed of 9 m/s with respect to the ground. The rotational speed of the platform after the balls leave is (in rad/s)
(a) 2
(b) 6
(c) 4
(d) 9
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. Angular momentum of the two balls about the axis: \( 2(0.05)(9)(0.25)=0.225 \) kg \( m^{2}/s \). The platform gets the same magnitude in the opposite sense.
2. \( I=\tfrac12(0.45)(0.5)^2=0.05625 \) kg \( m^{2} \) (the guns are massless).
3. \( \omega=\dfrac{0.225}{0.05625}=4 \) rad/s.

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Question 5: A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with angular velocity 10 rad/s about its own vertical axis. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are gently placed symmetrically on the disc, horizontally, touching each other at the axis (so each ring's centre is 0.2 m from the axis). Friction is large enough that the rings come to rest relative to the disc. The new angular velocity of the system (in rad/s) is
(a) 6
(b) 10
(c) 5
(d) 8
Show Answer & Explanation

Answer: (d) 8

Explanation:
1. \( I_{disc}=\tfrac12(50)(0.4)^2=4 \) kg \( m^{2} \).
2. Each ring about the axis: \( mr^2+md^2=6.25(0.04+0.04)=0.5 \), so two rings add 1.0 kg \( m^{2} \).
3. \( 4(10)=(4+1)\omega\Rightarrow\omega=8 \) rad/s.

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Question 6: A binary star consists of two stars A (mass 2.2 \( M_S \)) and B (mass 11 \( M_S \)), where \( M_S \) is the mass of the sun. They are separated by a distance d and rotate about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is
(a) 6
(b) 4
(c) 8
(d) 11
Show Answer & Explanation

Answer: (a) 6

Explanation:
1. Distances from the centre of mass: \( r_A=\dfrac{11d}{13.2} \), \( r_B=\dfrac{2.2d}{13.2} \).
2. Both have the same \( \omega \): \( L\propto mr^2 \). \( 2.2(11)^2=266.2 \), \( 11(2.2)^2=53.24 \) (common factor omitted).
3. Ratio \( =\dfrac{266.2+53.24}{53.24}=6 \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Question 7: A stone of mass m tied to the end of a string is whirled around in a horizontal circle (neglect gravity). The length of the string is reduced gradually keeping the angular momentum of the stone about the centre of the circle constant. The tension in the string is then \( T=Ar^n \), where A is a constant and r is the instantaneous radius. The value of n is
(a) -1
(b) -3
(c) -2
(d) 3
Show Answer & Explanation

Answer: (b) -3

Explanation:
1. \( L=mvr \) constant \( \Rightarrow v\propto\dfrac1r \).
2. \( T=\dfrac{mv^2}r\propto\dfrac1{r^3} \), so \( n = -3 \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Question 8: A smooth uniform rod of length L and mass M has two identical beads of negligible size, each of mass m, which can slide freely along the rod. Initially the two beads are at the centre of the rod and the system is rotating with angular velocity \( \omega_0 \) about an axis perpendicular to the rod and passing through its mid-point. There are no external forces. When the beads reach the ends of the rod, the angular velocity of the system is
(a) \( \dfrac{M\omega_0}{M+3m} \)
(b) \( \dfrac{M\omega_0}{M+2m} \)
(c) \( \dfrac{M\omega_0}{M+6m} \)
(d) \( \dfrac{2M\omega_0}{M+6m} \)
Show Answer & Explanation

Answer: (c) \( \dfrac{M\omega_0}{M+6m} \)

Explanation:
1. \( \dfrac{ML^2}{12}\omega_0=\left(\dfrac{ML^2}{12}+2m\dfrac{L^2}4\right)\omega \).
2. \( \omega=\dfrac{M\omega_0}{M+6m} \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Question 9: A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis through its centre and perpendicular to its plane with angular velocity \( \omega \). Another disc of the same dimensions but of mass M/4 is placed gently on the first disc coaxially. The angular velocity of the system now is
(a) \( \dfrac{2\omega}{\sqrt5} \)
(b) \( \dfrac{3\omega}{4} \)
(c) \( \dfrac{\omega}{2} \)
(d) \( \dfrac{4\omega}{5} \)
Show Answer & Explanation

Answer: (d) \( \dfrac{4\omega}{5} \)

Explanation:
1. \( I_1=\tfrac12MR^2 \), \( I_2=\tfrac12\dfrac M4R^2 \). Angular momentum is conserved.
2. \( \omega'=\dfrac{I_1\omega}{I_1+I_2}=\dfrac{\omega}{1+1/4}=\dfrac{4\omega}5 \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Question 10: A particle is projected at t = 0 from a point P on the ground with speed \( v_0 \) at an angle of \( 45^\circ \) to the horizontal. The magnitude and direction of its angular momentum about P at time \( t=\dfrac{v_0}{g} \) is
(a) \( \dfrac{mv_0^3}{2\sqrt2\,g} \), perpendicular to the plane of motion, into the plane (clockwise)
(b) \( \dfrac{mv_0^3}{2\sqrt2\,g} \), perpendicular to the plane of motion, out of the plane
(c) \( \dfrac{mv_0^3}{\sqrt2\,g} \), into the plane
(d) \( \dfrac{mv_0^3}{4\sqrt2\,g} \), into the plane
Show Answer & Explanation

Answer: (a) \( \dfrac{mv_0^3}{2\sqrt2\,g} \), perpendicular to the plane of motion, into the plane (clockwise)

Explanation:
1. At \( t=v_0/g \): \( x=\dfrac{v_0^2}{\sqrt2g} \), \( y=\dfrac{v_0^2}{\sqrt2g}-\dfrac{v_0^2}{2g} \), \( v_x=\dfrac{v_0}{\sqrt2} \), \( v_y=\dfrac{v_0}{\sqrt2}-v_0 \).
2. \( L_z=m(xv_y-yv_x)=\dfrac{mv_0^3}{g}\cdot\dfrac1{\sqrt2}\left[\left(\tfrac1{\sqrt2}-1\right)-\left(\tfrac1{\sqrt2}-\tfrac12\right)\right]=-\dfrac{mv_0^3}{2\sqrt2g} \).
3. The negative sign means clockwise, i.e. into the plane of motion.

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Question 11: A metal coin of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible mass, with AB along a diameter of the coin. The system is initially at rest. The constant torque that will make the system rotate about AB at 25 rotations per second in 5 s is close to
(a) \( 2.0\times10^{-5} \) N m
(b) \( 4.0\times10^{-6} \) N m
(c) \( 1.6\times10^{-5} \) N m
(d) \( 7.9\times10^{-6} \) N m
Show Answer & Explanation

Answer: (b) \( 4.0\times10^{-6} \) N m

Explanation:
1. \( I=\tfrac14mR^2=\tfrac14(5\times10^{-3})(10^{-2})^2=1.25\times10^{-7} \) kg \( m^{2} \).
2. \( \omega=2\pi(25)=157 \) rad/s, \( \alpha=\dfrac{157}5=31.4 \) rad/s².
3. \( \tau=I\alpha=1.25\times10^{-7}\times31.4\approx4.0\times10^{-6} N m \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 12: The moment of inertia of a body about a given axis is 1.5 kg m². Initially the body is at rest. To produce a rotational kinetic energy of 1200 J, an angular acceleration of 20 rad/s² must be applied about that axis for a duration of
(a) 5 s
(b) 3 s
(c) 2 s
(d) 2.5 s
Show Answer & Explanation

Answer: (c) 2 s

Explanation:
1. \( \tfrac12I\omega^2=1200\Rightarrow\omega^2=1600\Rightarrow\omega=40 \) rad/s.
2. \( t=\dfrac\omega\alpha=\dfrac{40}{20}=2 s \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 13: A ring of radius R, a solid cylinder of radius R/2 and a solid sphere of radius R/4 are made to roll up (without slipping) the same inclined plane. In each case the speed of the centre of mass at the bottom of the incline is the same. The ratio of the maximum heights they climb (ring : cylinder : sphere) is
(a) 10 : 15 : 7
(b) 4 : 3 : 2
(c) 14 : 15 : 20
(d) 20 : 15 : 14
Show Answer & Explanation

Answer: (d) 20 : 15 : 14

Explanation:
1. Energy: \( \tfrac12mv^2\left(1+\dfrac I{mr^2}\right)=mgh\Rightarrow h=\dfrac{v^2}{2g}\left(1+\dfrac I{mr^2}\right) \), independent of radius.
2. Factors: ring \( 1+1=2 \), solid cylinder \( 1+\tfrac12=\tfrac32 \), solid sphere \( 1+\tfrac25=\tfrac75 \).
3. Ratio \( 2:\tfrac32:\tfrac75=20:15:14 \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 14: A solid sphere and a solid cylinder of identical radii approach an incline with the same linear velocity. Both roll without slipping throughout. The two climb maximum heights \( h_{sph} \) and \( h_{cyl} \) on the incline. The ratio \( h_{sph}/h_{cyl} \) is
(a) \( \dfrac{14}{15} \)
(b) \( \dfrac{2}{\sqrt5} \)
(c) \( 1 \)
(d) \( \dfrac{4}{5} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{14}{15} \)

Explanation:
1. \( h=\dfrac{v^2}{2g}\left(1+\dfrac I{mr^2}\right) \). Sphere: \( 1+\tfrac25=\tfrac75 \); cylinder: \( 1+\tfrac12=\tfrac32 \).
2. \( \dfrac{h_{sph}}{h_{cyl}}=\dfrac{7/5}{3/2}=\dfrac{14}{15} \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 15: A string is wound around a hollow cylinder of mass 5 kg and radius 0.5 m. The string is pulled with a horizontal force of 40 N from the top of the cylinder, and the cylinder rolls without slipping on a horizontal surface (neglect the mass and thickness of the string). The angular acceleration of the cylinder is
(a) 10 rad/s²
(b) 16 rad/s²
(c) 20 rad/s²
(d) 12 rad/s²
Show Answer & Explanation

Answer: (b) 16 rad/s²

Explanation:
1. Take torques about the contact point: \( F(2R)=(I+mR^2)\alpha=2mR^2\alpha \).
2. \( \alpha=\dfrac{F}{mR}=\dfrac{40}{5\times0.5}=16 \) rad/s².

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 16: A homogeneous solid cylindrical roller of radius R and mass m is pulled on a cricket pitch by a horizontal force F applied at its axis. Assuming it rolls without slipping, its angular acceleration is
(a) \( \dfrac{F}{2mR} \)
(b) \( \dfrac{3F}{2mR} \)
(c) \( \dfrac{2F}{3mR} \)
(d) \( \dfrac{F}{3mR} \)
Show Answer & Explanation

Answer: (c) \( \dfrac{2F}{3mR} \)

Explanation:
1. \( F-f=ma \) and \( fR=\tfrac12mR^2\alpha \) with \( a=\alpha R \), so \( f=\tfrac12ma \).
2. \( F=\tfrac32ma\Rightarrow\alpha=\dfrac aR=\dfrac{2F}{3mR} \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 17: A mass m is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R, which can rotate freely about its fixed horizontal axis. If the string does not slip on the cylinder, the acceleration with which the mass falls on release is
(a) 2g/3
(b) 5g/6
(c) g
(d) g/2
Show Answer & Explanation

Answer: (d) g/2

Explanation:
1. \( mg-T=ma \) and \( TR=mR^2\alpha=mRa \), so \( T=ma \).
2. \( mg=2ma\Rightarrow a=\dfrac g2 \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 18: Two solid cylinders P and Q of the same mass and the same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement is correct?
(a) Cylinder Q reaches the ground with larger angular speed
(b) Both cylinders reach the ground at the same time
(c) Cylinder P has larger linear acceleration than cylinder Q
(d) Both cylinders reach the ground with the same translational kinetic energy
Show Answer & Explanation

Answer: (a) Cylinder Q reaches the ground with larger angular speed

Explanation:
1. \( a=\dfrac{g\sin\theta}{1+I/mR^2} \): \( Q \) has the smaller I, so it has the larger acceleration, reaches first and has the larger speed.
2. Hence \( Q \) has the larger \( \omega=v/R \). Translational KE is \( \dfrac{mgh}{1+I/mR^2} \), which is larger for \( Q \), so the other statements are false.

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 19: A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of \( \dfrac{3v^2}{4g} \) with respect to the initial position. The object is a
(a) ring
(b) disc
(c) solid sphere
(d) hollow sphere
Show Answer & Explanation

Answer: (b) disc

Explanation:
1. \( \tfrac12mv^2\left(1+\dfrac I{mr^2}\right)=mgh=\dfrac{3mv^2}4 \).
2. \( 1+\dfrac I{mr^2}=\dfrac32\Rightarrow\dfrac I{mr^2}=\dfrac12 \), which is a disc (or solid cylinder).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 20: A disc rolls without slipping on a horizontal surface. C is its centre, and Q and P are two points on the vertical diameter, Q above C and P below C, each at the same distance from C. If \( v_P \), \( v_Q \) and \( v_C \) are the speeds of P, Q and C respectively, then
(a) \( v_Q<v_C<v_P \)
(b) \( v_Q=v_P,\ v_C=\tfrac12v_P \)
(c) \( v_Q>v_C>v_P \)
(d) \( v_Q<v_C>v_P \)
Show Answer & Explanation

Answer: (c) \( v_Q>v_C>v_P \)

Explanation:
1. Rolling is pure rotation about the contact point, so speed is proportional to distance from it.
2. \( Q \) is farthest, \( C \) is at \( R \) and \( P \) is nearest: \( v_Q>v_C>v_P \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 21: A cylinder rolls up an inclined plane, reaches some height and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are
(a) up the incline while ascending and down the incline while descending
(b) down the incline while ascending and up the incline while descending
(c) down the incline while ascending as well as descending
(d) up the incline while ascending as well as descending
Show Answer & Explanation

Answer: (d) up the incline while ascending as well as descending

Explanation:
1. In both cases the centre of mass accelerates down the incline, and the cylinder's angular acceleration must match \( a/R \).
2. The torque needed for this comes from friction acting up the incline in both ascent and descent.

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 22:
Assertion (A): Two cylinders, one hollow (metal) and the other solid (wood), with the same mass and identical dimensions, are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first.
Reason (R): By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. Acceleration \( a=\dfrac{g\sin\theta}{1+I/mR^2} \) is larger for the solid cylinder (smaller \( I \)), so the solid cylinder reaches first: Statement I is false.
2. Both lose the same potential energy \( mgh \), so their total KE at the bottom is the same: Statement II is true.

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 23: A uniform thin cylindrical disk of mass M and radius R is attached to two identical massless springs, each of spring constant k, fixed to a wall. The springs are attached to the axle of the disk symmetrically on either side of its centre, along the x-direction. The unstretched length of each spring is L. The disk is initially at its equilibrium position with its centre of mass at distance L from the wall. The disk rolls without slipping on the floor with initial velocity \( v_0\hat i \) and friction coefficient \( \mu \). The net external force acting on the disk when its centre of mass is at displacement x from the equilibrium position is
(a) \( -kx \)
(b) \( \dfrac{-4kx}{3} \)
(c) \( -2kx \)
(d) \( \dfrac{-2kx}{3} \)
Show Answer & Explanation

Answer: (b) \( \dfrac{-4kx}{3} \)

Explanation:
1. Spring force \( -2kx \) plus friction \( f \): \( Ma=-2kx+f \).
2. Rolling: \( fR=\tfrac12MR^2\cdot\dfrac aR\) with the friction opposing the rotation sense, so \( f=-\tfrac12Ma \).
3. \( Ma=-2kx-\tfrac12Ma\Rightarrow Ma=-\dfrac{4kx}3 \). This is the net external force.

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 24: For the disk rolling between two springs (mass M, radius R, each spring constant k, attached to the axle on either side; the net force on the centre of mass is \( -\tfrac43kx \) at displacement x), the centre of mass of the disk performs simple harmonic motion with angular frequency \( \omega \) equal to
(a) \( \sqrt{\dfrac kM} \)
(b) \( \sqrt{\dfrac{2k}M} \)
(c) \( \sqrt{\dfrac{4k}{3M}} \)
(d) \( \sqrt{\dfrac{2k}{3M}} \)
Show Answer & Explanation

Answer: (c) \( \sqrt{\dfrac{4k}{3M}} \)

Explanation:
1. \( Ma=-\dfrac43kx\Rightarrow a=-\dfrac{4k}{3M}x \).
2. \( \omega=\sqrt{\dfrac{4k}{3M}} \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 25: A uniform disk of mass M and radius R rolls without slipping on a floor while attached to two identical springs of constant k on either side of its axle. The centre of mass performs SHM with \( \omega=\sqrt{\dfrac{4k}{3M}} \). The coefficient of friction between disk and floor is \( \mu \). The disk is at its equilibrium position with velocity \( v_0 \). The maximum value of \( v_0 \) for which the disk will roll without slipping is
(a) \( \mu g\sqrt{\dfrac Mk} \)
(b) \( \mu g\sqrt{\dfrac M{2k}} \)
(c) \( \mu g\sqrt{\dfrac{5M}{2k}} \)
(d) \( \mu g\sqrt{\dfrac{3M}{k}} \)
Show Answer & Explanation

Answer: (d) \( \mu g\sqrt{\dfrac{3M}{k}} \)

Explanation:
1. Amplitude \( A=\dfrac{v_0}\omega \). Maximum acceleration \( a_{max}=\omega^2A=\omega v_0 \).
2. Friction needed: \( f=\tfrac12Ma_{max}\le\mu Mg\Rightarrow v_0\le\dfrac{2\mu g}{\omega} \).
3. With \( \omega=\sqrt{\dfrac{4k}{3M}} \): \( v_0\le\mu g\sqrt{\dfrac{3M}{k}} \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 26: A system has a ring of outer radius 3R rolling clockwise without slipping with angular speed \( \omega \) on a horizontal surface (the x-axis points along the surface, z points up, and the ring moves along +x). Inside it, separated by frictionless bearings, an inner disc of radius 2R rotates anticlockwise with angular speed \( \omega/2 \) about the same centre O. A point P on the inner disc is at distance R from O, with OP making \( 30^\circ \) with the horizontal x-axis, above it on the +x side. Which statements are correct? (a) The point O has a linear velocity \( 3R\omega\hat i \). (b) The point P has a linear velocity \( \dfrac{11}4R\omega\hat i+\dfrac{\sqrt3}4R\omega\hat k \). (c) The point P has a linear velocity \( \dfrac{13}4R\omega\hat i-\dfrac{\sqrt3}4R\omega\hat k \). (d) The point P has a linear velocity \( \left(3-\dfrac{\sqrt3}4\right)R\omega\hat i+\dfrac14R\omega\hat k \).
(a) (a) and (b) only
(b) (a) and (c) only
(c) (b) and (d) only
(d) (c) and (d) only
Show Answer & Explanation

Answer: (a) (a) and (b) only

Explanation:
1. The ring rolls with speed \( 3R\omega \), so the common centre \( O \) moves with \( 3R\omega\hat i \): \( (a) \) true.
2. Relative to \( O \), point \( P \) at \( \left(\tfrac{\sqrt3}2R,\tfrac R2\right) \) moves, for anticlockwise \( \omega/2 \), with \( \dfrac\omega2(-y,x)=\left(-\dfrac{R\omega}4,\dfrac{\sqrt3R\omega}4\right) \).
3. Adding \( 3R\omega\hat i \): \( \vec v_P=\dfrac{11}4R\omega\hat i+\dfrac{\sqrt3}4R\omega\hat k \): \( (b) \) true, \( (c) \) and \( (d) \) false.

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 27: A sphere is rolling without slipping on a fixed horizontal plane. A is the point of contact, B is the centre of the sphere and C is its topmost point. Which of the following are correct? (a) \( \vec v_C-\vec v_A=2(\vec v_B-\vec v_C) \) (b) \( \vec v_C-\vec v_B=\vec v_B-\vec v_A \) (c) \( |\vec v_C-\vec v_A|=2|\vec v_B-\vec v_C| \) (d) \( |\vec v_C-\vec v_A|=4|\vec v_B| \)
(a) (a) and (b) only
(b) (b) and (c) only
(c) (c) and (d) only
(d) (a) and (d) only
Show Answer & Explanation

Answer: (b) (b) and (c) only

Explanation:
1. With centre velocity \( \vec v \): \( \vec v_A=0 \), \( \vec v_B=\vec v \), \( \vec v_C=2\vec v \).
2. \( (a) 2\vec v \) vs \( -2\vec v \): false. \( (b) \vec v=\vec v \): true. \( (c) 2v=2v \): true. \( (d) 2v \) vs \( 4v \): false.

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 28: A solid sphere is in pure rolling motion down an inclined surface of inclination \( \theta \). Which statements are correct? (a) The frictional force on the sphere is \( f=\mu mg\cos\theta \). (b) f is dissipative. (c) Friction will increase its angular velocity and decrease its linear velocity (compared with frictionless sliding). (d) If \( \theta \) decreases, friction will decrease.
(a) (a) and (b) only
(b) (a) and (c) only
(c) (c) and (d) only
(d) (b) and (d) only
Show Answer & Explanation

Answer: (c) (c) and (d) only

Explanation:
1. For pure rolling, static friction is \( f=\tfrac27mg\sin\theta \), not \( \mu mg\cos\theta \), and it does no net work, so it is not dissipative: \( (a), (b) \) false.
2. Friction up the slope reduces the linear acceleration and provides the torque that spins the sphere up: \( (c) \) true. \( f\propto\sin\theta \) decreases with \( \theta \): \( (d) \) true.

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 29: A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle of \( 60^\circ \) with the horizontal. They start to roll without slipping at the same instant along the shortest path. If the time difference between their reaching the ground is \( (2-\sqrt3)/\sqrt{10} \) s, the height of the top of the inclined plane is (g = 10 m/s²)
(a) 0.5 m
(b) 1.0 m
(c) 1.5 m
(d) 0.75 m
Show Answer & Explanation

Answer: (d) 0.75 m

Explanation:
1. Along the plane of length \( L \): \( a_{ring}=\dfrac{g\sin\theta}2 \), \( a_{disc}=\dfrac{2g\sin\theta}3 \).
2. \( t=\sqrt{2L/a} \): \( t_{ring}=2\sqrt{\dfrac L{g\sin\theta}} \), \( t_{disc}=\sqrt3\sqrt{\dfrac L{g\sin\theta}} \). Difference \( =(2-\sqrt3)\sqrt{\dfrac L{g\sin\theta}} \).
3. So \( \dfrac L{g\sin\theta}=\dfrac1{10} \Rightarrow L=\sin60^\circ=0.866 m \) and \( h=L\sin60^\circ=0.75 m \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 30: A uniform disc of mass m and radius R is rolling up a rough inclined plane which makes an angle of \( 30^\circ \) with the horizontal. The only forces acting on the disc are gravitational and frictional, and the disc rolls without slipping. The magnitude and direction of the frictional force on the disc are
(a) mg/6, up the incline
(b) mg/6, down the incline
(c) mg/3, up the incline
(d) mg/2, down the incline
Show Answer & Explanation

Answer: (a) mg/6, up the incline

Explanation:
1. \( mg\sin\theta-f=ma \) (deceleration \( a \)) and \( fR=\tfrac12mR^2\dfrac aR\Rightarrow f=\tfrac12ma \).
2. \( mg\sin\theta=\tfrac32ma\Rightarrow f=\dfrac{mg\sin\theta}3=\dfrac{mg}6 \), directed up the incline.

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Question 31: A cylinder of mass M and radius R rests on a horizontal platform (parallel to the x-y plane) with its axis along the Y-axis and free to rotate about its axis. The platform moves in the x-direction as \( x=A\cos\omega t \), and there is no slipping between the cylinder and the platform. The maximum torque acting on the cylinder during its motion is
(a) \( \dfrac12MRA\omega^2 \)
(b) \( \dfrac13MRA\omega^2 \)
(c) \( \dfrac23MRA\omega^2 \)
(d) \( MRA\omega^2 \)
Show Answer & Explanation

Answer: (b) \( \dfrac13MRA\omega^2 \)

Explanation:
1. Friction \( f \) gives \( a_{cm}=f/M \) and \( \alpha=\dfrac{fR}{\tfrac12MR^2}=\dfrac{2f}{MR} \). No slipping: platform acceleration \( =a_{cm}+\alpha R=\dfrac{3f}M \).
2. \( f=\dfrac M3a_{platform} \), maximum \( \dfrac{MA\omega^2}3 \). Torque \( =fR=\dfrac13MRA\omega^2 \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

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