JEE Physics System Of Particles and Rotational Motion MCQs Set 42

Answer: (d) 20 rad/s

Explanation:
1. Relative speed of the end with respect to the centre is \( 20-10=10 \) m/s at a distance 0.5 m.
2. \( \omega=\dfrac{10}{0.5}=20 \) rad/s.

Teacher's Note:
1. Subtract the centre-of-mass velocity.
2. The end is 0.5 m from the centre.

Answer: (a) The hollow sphere reaches the bottom with lesser speed

Explanation:
1. Pure rolling: \( v^2=\dfrac{2gh}{1+k^2/R^2} \). Solid sphere k²/R² = 2/5; hollow sphere 2/3.
2. So the solid sphere is faster, and the hollow sphere reaches the bottom with a smaller speed (and takes more time).
3. Kinetic energy at the bottom is mgh for both; momentum mv differs.

Teacher's Note:
1. Greater moment of inertia means more of the energy goes into rotation.
2. Total energy at the bottom is the same for both.

Answer: (d) it will continue pure rolling

Explanation:
1. In the car's frame the effective gravity is \( g_{eff}=\sqrt{g^2+a^2}=g/\cos\theta \), directed perpendicular to the incline.
2. So there is no force component along the incline: no net force, no friction needed and no torque. The sphere keeps rolling with constant speed.

Teacher's Note:
1. Find the direction of the effective gravity in the accelerating frame.
2. No force along the plane means constant velocity.

Answer: (b) it will move forward before pure rolling begins

Explanation:
1. Friction opposes the sliding of the contact point and acts backward, reducing v while producing forward spin.
2. The body keeps moving forward until \( v=\omega R \), then rolls.

Teacher's Note:
1. The linear velocity is reduced but not reversed.
2. Pure rolling begins when the contact point stops slipping.

Answer: (b) always forward

Explanation:
1. The contact point slides backward (opposite to the top), so kinetic friction acts forward.
2. The body accelerates forward while its spin falls, until pure rolling; it never moves backward.

Teacher's Note:
1. Friction is forward because the contact point slides backward.
2. The centre accelerates forward continuously.

Answer: (a) a circular disc

Explanation:
1. Energy: \( \tfrac12mv^2(1+k^2/R^2)=mgh=\dfrac{3mv^2}{4} \Rightarrow 1+\dfrac{k^2}{R^2}=\dfrac32 \).
2. \( k^2/R^2=\tfrac12 \), which is the disc.

Teacher's Note:
1. Each shape has a characteristic ratio of k squared to R squared.
2. Disc 1/2, sphere 2/5, ring 1, shell 2/3.

Answer: (c) \( \dfrac{V}{3R} \)

Explanation:
1. The centre of mass of the system is at \( R/2 \) from the loop's centre towards the bullet. Angular momentum about it before: \( mV\cdot\dfrac R2 \).
2. Moment of inertia about it: \( mR^2+m\dfrac{R^2}{4}+m\dfrac{R^2}{4}=\dfrac32mR^2 \).
3. \( \omega=\dfrac{mVR/2}{3mR^2/2}=\dfrac{V}{3R} \).

Teacher's Note:
1. Take angular momentum about the centre of mass of the combined system.
2. The bullet is at R/2 from that point, and so is the centre of the loop.

Answer: (c) \( \dfrac23mg \)

Explanation:
1. At the point of tilting, the normal force acts at the edge. Torque about that edge: \( F\cdot\dfrac{3a}{4}=mg\cdot\dfrac a2 \).
2. \( F=\dfrac23mg \).

Teacher's Note:
1. Take torques about the pivoting edge.
2. The weight's arm is half the side.

Answer: (b) \( \cos\theta\le\dfrac ba \)

Explanation:
1. The vertical through the centre of mass must pass inside the base.
2. The centre is at the mid-point of the parallelogram, horizontally displaced from the middle of the base by \( \dfrac{a\cos\theta}{2} \). We need \( \dfrac{a\cos\theta}{2}\le\dfrac b2 \).
3. \( \cos\theta\le b/a \).

Teacher's Note:
1. Stable equilibrium needs the centre of mass to lie above the base.
2. Use the horizontal projection of the slanted side.

Answer: (a) \( \dfrac{mg}{\sqrt3} \)

Explanation:
1. The prism tips about the lower vertex on the side opposite the push. The height of the apex is \( \dfrac{\sqrt3a}{2} \); the horizontal distance of the centre of mass from that vertex is a/2.
2. \( F\cdot\dfrac{\sqrt3a}{2}=mg\cdot\dfrac a2\Rightarrow F=\dfrac{mg}{\sqrt3} \).

Teacher's Note:
1. Friction is assumed large enough, so the coefficient does not appear.
2. Take torques about the corner that stays on the ground.

Answer: (d) \( \sqrt{\dfrac76}\,r \)

Explanation:
1. Top ring: axis along a diameter, \( I=\tfrac12mr^2 \).
2. Each lower ring: the axis is parallel to a diameter at distance r, \( I=\tfrac12mr^2+mr^2=\tfrac32mr^2 \).
3. Total \( I=\tfrac12+3=\tfrac72mr^2=3mk^2\Rightarrow k=\sqrt{7/6}\,r \).

Teacher's Note:
1. Use the parallel axes theorem for each lower ring.
2. Radius of gyration is the square root of I over the total mass.

Answer: (d) \( \dfrac M2(3R_2^2+R_1^2) \)

Explanation:
1. About the symmetry axis: \( I_{cm}=\dfrac M2(R_1^2+R_2^2) \).
2. Parallel axes with \( d=R_2 \): \( I=\dfrac M2(R_1^2+R_2^2)+MR_2^2=\dfrac M2(3R_2^2+R_1^2) \).

Teacher's Note:
1. The tangential axis is at distance R2 from the centre.
2. Use the parallel axes theorem.

Answer: (a) \( \sqrt{\dfrac{2g}{L}} \)

Explanation:
1. In the vertical position the torque of the weight is zero, so \( \alpha=0 \) and the centre of mass has only a centripetal acceleration \( \omega^2L/2 \) downwards.
2. \( mg-N=m\omega^2\dfrac L2 \). For \( N=0 \): \( \omega^2=\dfrac{2g}{L} \).

Teacher's Note:
1. There is no tangential acceleration at the vertical position.
2. The weight alone must provide the centripetal force.

Answer: (a) \( \dfrac{5mg}{8},\ \dfrac{3g}{8\ell} \)

Explanation:
1. Block: \( mg-T=ma \), and its acceleration equals that of B: \( a=\alpha\ell \).
2. Rod about A: \( T\ell-mg\dfrac\ell2=\dfrac{m\ell^2}{3}\alpha \).
3. Substituting \( T=mg-m\alpha\ell \): \( \dfrac{mg}{2}-m\alpha\ell=\dfrac{m\alpha\ell}{3}\Rightarrow\alpha=\dfrac{3g}{8\ell} \) and \( T=\dfrac{5mg}{8} \).

Teacher's Note:
1. The block and the end of the rod have the same acceleration.
2. Take torque about the hinge so the hinge force drops out.

Answer: (c) \( \dfrac\ell2\left(1+\dfrac\pi6\right) \)

Explanation:
1. Let \( J=F\Delta t \). \( v_{cm}=J/m \); \( J\dfrac\ell2=\dfrac{m\ell^2}{12}\omega\Rightarrow\omega=\dfrac{6J}{m\ell} \).
2. Rotation by \( \pi/2 \) takes \( t=\dfrac{\pi}{2\omega}=\dfrac{\pi m\ell}{12J} \), the centre moving \( v_{cm}t=\dfrac{\pi\ell}{12} \).
3. End A has swung to the +x side: \( x_A=\dfrac{\pi\ell}{12}+\dfrac\ell2=\dfrac\ell2\left(1+\dfrac\pi6\right) \).

Teacher's Note:
1. Treat the translation of the centre of mass and the spin separately.
2. The end struck moves forward relative to the centre.

Answer: (a) \( 2\mu g \)

Explanation:
1. Kinetic friction \( \mu mg \) acts on each in opposite directions.
2. Sphere: \( a_s=\mu g \) one way; plank: \( a_p=\mu g \) the other way.
3. Relative acceleration \( =2\mu g \).

Teacher's Note:
1. The friction on the plank is the reaction of the friction on the sphere.
2. Add the accelerations when the two move in opposite senses.

Answer: (d) \( \dfrac14\sqrt{\dfrac{40g\ell}{19}},\ \dfrac34\sqrt{\dfrac{40g\ell}{19}},\ \sqrt{\dfrac{40g}{19\ell}} \)

Explanation:
1. The right mass (at 3l/4) falls by 3l/4 and the left mass (at l/4) rises by l/4: energy gain \( 2mg\dfrac{3\ell}{4}-mg\dfrac\ell4=\dfrac54mg\ell \).
2. \( I=m\dfrac{\ell^2}{16}+2m\dfrac{9\ell^2}{16}=\dfrac{19}{16}m\ell^2 \). \( \dfrac54mg\ell=\tfrac12\cdot\dfrac{19}{16}m\ell^2\omega^2\Rightarrow\omega^2=\dfrac{40g}{19\ell} \).
3. Speeds: left \( \omega\ell/4 \), right \( 3\omega\ell/4 \).

Teacher's Note:
1. The rod has negligible mass, so only the two particles matter.
2. Speeds are the angular velocity times the distances from the pivot.

Answer: (c) \( \dfrac{mgh}{7} \)

Explanation:
1. A falls through h/2 while rolling: total energy \( \tfrac12mv^2\cdot\dfrac75=mg\dfrac h2 \), so \( \tfrac12mv^2=\dfrac{5mgh}{14} \) and the spin energy is \( \dfrac{2}{7}\cdot\dfrac{mgh}{2}=\dfrac{mgh}{7} \).
2. Elastic collision of equal masses with no friction between the balls: velocities along the line of centres are exchanged. A loses all translational speed; its spin is unchanged.
3. KE of A afterwards \( =\dfrac{mgh}{7} \).

Teacher's Note:
1. Smooth contact between the balls means the spin of A does not change.
2. Equal masses in an elastic collision exchange velocities.

Answer: (b) \( \dfrac{g}{8r} \)

Explanation:
1. Take torques about the contact point: the weight of the block has arm r, the ring's weight passes through the line of the centre (zero torque).
2. Moment of inertia about the contact point: ring \( 3mr^2+3mr^2=6mr^2 \), block at distance \( \sqrt2r \): \( 2mr^2 \). Total \( 8mr^2 \).
3. \( \alpha=\dfrac{mgr}{8mr^2}=\dfrac{g}{8r} \).

Teacher's Note:
1. Use the contact point as the instantaneous axis.
2. Check by the Lagrangian method: the same result follows.

Answer: (a) \( \dfrac{V_0}{4} \)

Explanation:
1. Angular momentum about the contact point is conserved: \( mV_0R-mR^2\dfrac{V_0}{2R}=\dfrac12mV_0R \).
2. Pure rolling: \( mVR+mR^2\dfrac VR=2mVR \). So \( V=\dfrac{V_0}{4} \).

Teacher's Note:
1. Friction acts at the contact point, so angular momentum about it is conserved.
2. Backspin is opposite to the rolling sense, giving a negative contribution.

Answer: (d) \( \dfrac{gb}{\ell} \)

Explanation:
1. In the truck's frame a pseudo force ma acts at the centre of mass, at height \( \ell/2 \).
2. About the lower edge: toppling begins when \( ma\dfrac\ell2=mg\dfrac b2 \).
3. \( a_{max}=\dfrac{gb}{\ell} \).

Teacher's Note:
1. Take torque about the edge that stays on the truck floor.
2. A tall narrow box topples sooner.

Answer: (c) \( \dfrac{m_1m_2}{m_1+m_2}r_0^2 \)

Explanation:
1. Distances from the centre of mass: \( \dfrac{m_2r_0}{m_1+m_2} \) and \( \dfrac{m_1r_0}{m_1+m_2} \).
2. \( I=m_1\dfrac{m_2^2r_0^2}{(m_1+m_2)^2}+m_2\dfrac{m_1^2r_0^2}{(m_1+m_2)^2}=\dfrac{m_1m_2}{m_1+m_2}r_0^2 \).

Teacher's Note:
1. This is the reduced mass times the square of the separation.
2. It is the same for any orientation of the axis perpendicular to the rod.

Answer: (b) energy, \( g/3 \)

Explanation:
1. Net driving weight \( =(4-2)g \), total mass \( =6 \) kg: \( a=\dfrac{2g}{6}=\dfrac g3 \).
2. The pulley is smooth and the string inextensible, so mechanical energy is conserved. Linear momentum and angular momentum about the pulley axis change because gravity exerts a net torque.

Teacher's Note:
1. Only conservative forces and constraint forces doing no work act.
2. The net external torque about the pulley axis is not zero.

Answer: (a) 16

Explanation:
1. First 20 s: \( \theta_1=\tfrac12(2)(20)^2=400 \) rad, final \( \omega=40 \) rad/s.
2. Next 20 s: \( \theta_2=40\times20=800 \) rad. Last 20 s: \( \theta_3=\dfrac{40}{2}\times20=400 \) rad.
3. Total 1600 rad, so n = 16.

Teacher's Note:
1. Break the motion into three phases.
2. The deceleration phase has average speed 20 rad/s.

Answer: (a) 2

Explanation:
1. Each rod about the centroid axis: \( \dfrac{m\ell^2}{12}+m\left(\dfrac{\ell}{2\sqrt3}\right)^2=\dfrac{m\ell^2}{6} \); three rods \( \dfrac{m\ell^2}{2} \).
2. Shift to a corner (distance \( \ell/\sqrt3 \)): \( I=\dfrac{m\ell^2}{2}+3m\dfrac{\ell^2}{3}=\dfrac32m\ell^2 \).
3. \( k^2=\dfrac{I}{3m}=\dfrac{\ell^2}{2} \), so n = 2.

Teacher's Note:
1. The distance from the centroid to a side is l over 2 root 3.
2. Use the parallel axes theorem twice.