JEE Physics System Of Particles and Rotational Motion MCQs Set 36

Answer: (a) \( \frac{mgl}{2} \)

Explanation:
1. The centre of mass of the frame is \( l/2 \) from the axis AD, so after 90 degrees it is lower by \( l/2 \).
2. \( W=mg\frac l2 \)

Teacher's Note:
1. Work done by gravity equals weight times the fall of the centre of mass.

Answer: (c) 3.9 ft/s

Explanation:
1. In the car frame the pseudo force \( ma \) acts at the centre of the door, with lever arm \( \frac w2\cos\phi \) as the door closes through 90 degrees.
2. Work done \( =ma\frac w2 \) equals \( \frac12\cdot\frac{mw^2}{3}\omega^2 \), so \( \omega^2=\frac{3a}{w} \).
3. \( v=\omega w=\sqrt{3aw}=\sqrt{3\times2\times2.5}=\sqrt{15}\approx3.9\ \text{ft/s} \)

Teacher's Note:
1. Use work done by the pseudo force in the accelerating frame.

Answer: (b) 2Mg

Explanation:
1. \( \omega^2=\frac{3g}{L} \). The part beyond L/3 has mass \( \frac{2M}3 \) and its centre is at \( \frac{2L}3 \) from the hinge.
2. \( T=\frac{2M}{3}\left(g+\omega^2\frac{2L}{3}\right)=\frac{2M}3(g+2g)=2Mg \)

Teacher's Note:
1. Tension equals weight plus centripetal force of the part of the rod beyond the point.

Answer: (b) 1.6 N

Explanation:
1. Block (decelerating, up the plane): \( mg\sin30^\circ-T=ma \). Drum: \( TR=\frac12MR^2\frac aR\Rightarrow T=\frac12Ma=a \) (M = 2 kg).
2. So \( 2.45-T=0.5T\Rightarrow T=1.63\ \text{N} \)

Teacher's Note:
1. The drum's rotational inertia acts like an extra mass M/2 for the block.

Answer: (b) 1.2 m

Explanation:
1. Deceleration \( a=\frac{mg\sin\theta}{m+M/2}=\frac{2.45}{1.5}=1.63\ \text{m/s}^2 \); speed \( v=\omega R=2\ \text{m/s} \).
2. \( s=\frac{v^2}{2a}=\frac{4}{3.27}\approx1.2\ \text{m} \)

Teacher's Note:
1. Use the speed of the string and the constant deceleration.

Answer: (b) 30 degrees clockwise from east (30 degrees south of east)

Explanation:
1. Zero total angular momentum: \( mR^2\omega_m+\frac12(22m)R^2\omega_d=0\Rightarrow\omega_d=-\frac{\omega_m}{11} \).
2. Relative angular speed \( \omega_m-\omega_d=\frac{12}{11}\omega_m \), so when the relative angle is \( 2\pi \), the man has turned \( \frac{11}{12}\times360^\circ=330^\circ \) with respect to the ground.
3. That is 30 degrees short of A, i.e. 30 degrees clockwise from east.

Teacher's Note:
1. The disc rotates oppositely, so the man covers less than a full turn relative to the ground.

Answer: (b) 3.33 m/s

Explanation:
1. The string jerk is impulsive; angular momentum about the pulley axis is conserved: \( mvr=mv'r+\frac12mr^2\frac{v'}{r} \).
2. \( v=\frac32v'\Rightarrow v'=\frac{10}{3}\approx3.33\ \text{m/s} \)

Teacher's Note:
1. Use angular momentum about the pulley axis during the sudden jerk.

Answer: (b) 144 degrees anticlockwise

Explanation:
1. Zero angular momentum: \( 2(75)R^2\omega_m+\frac12(450)R^2\omega_d=0\Rightarrow\omega_d=-\frac23\omega_m \).
2. Relative angle \( 2\pi \) is covered when \( \omega_m-\omega_d=\frac53\omega_m \) acts: the disc turns \( \frac{2/3}{5/3}\times360^\circ=144^\circ \) opposite to the men (anticlockwise).

Teacher's Note:
1. The disc recoils opposite to the men by a fraction of their relative turn.

Answer: (c) 75 J

Explanation:
1. \( J L=I\omega\Rightarrow\omega=\frac{10\times1}{\frac13\times2\times1}=15\ \text{rad/s} \).
2. \( K=\frac12I\omega^2=\frac12\times\frac23\times225=75\ \text{J} \)

Teacher's Note:
1. Angular impulse about the hinge equals the gain in angular momentum.

Answer: (a) mv/4

Explanation:
1. Angular momentum about A: \( mv\frac L2=\frac{mL^2}3\omega\Rightarrow\omega=\frac{3v}{2L} \); the centre of the rod moves at \( \frac{\omega L}2=\frac{3v}4 \).
2. Impulse from the particle is \( mv \). Linear impulse: \( mv+J_A=m\frac{3v}{4}\Rightarrow J_A=-\frac{mv}4 \), magnitude mv/4.

Teacher's Note:
1. Apply linear and angular impulse equations to the rod.

Answer: (b) 64.5

Explanation:
1. Distance \( =\frac{27}{2}\times9=121.5\ \text{m} \).
2. \( N=\frac{121.5}{\pi\times0.6}=64.5 \)

Teacher's Note:
1. Revolutions equal the distance covered divided by the tyre circumference.

Answer: (a) \( \frac{15}{14} \) and \( \frac43 \)

Explanation:
1. \( a=\frac{g\sin\theta}{1+k} \) is independent of radius: sphere \( \frac57g\sin\theta \), disc \( \frac23g\sin\theta \), ring \( \frac12g\sin\theta \).
2. \( \frac ab=\frac{5/7}{2/3}=\frac{15}{14},\ \frac bc=\frac{2/3}{1/2}=\frac43 \)

Teacher's Note:
1. Acceleration down an incline depends only on the shape factor k.

Answer: (a) \( \frac{ma}{2} \) up the plane

Explanation:
1. Torque about the centre: \( fR=\frac12mR^2\frac aR\Rightarrow f=\frac{ma}2 \), directed up the plane.

Teacher's Note:
1. Friction supplies the torque that spins up the disc.

Answer: (a) \( \frac{13}{16}Mv^2 \)

Explanation:
1. \( I=\frac12M(R^2+4R^2)=\frac52MR^2 \); \( \omega=\frac{v}{2R} \).
2. \( K=\frac12Mv^2+\frac12\cdot\frac52MR^2\cdot\frac{v^2}{4R^2}=\frac12Mv^2+\frac{5}{16}Mv^2=\frac{13}{16}Mv^2 \)

Teacher's Note:
1. Moment of inertia of a thick cylinder is half M times the sum of the squared radii.

Answer: (c) 4R

Explanation:
1. At the top the particle has speed \( 2v \) and acceleration \( \frac{v^2}{R} \) directed towards the centre (downward).
2. \( \rho=\frac{(2v)^2}{v^2/R}=4R \)

Teacher's Note:
1. Radius of curvature equals speed squared divided by normal acceleration.

Answer: (b) \( \frac{Ml^2}{18} \)

Explanation:
1. \( M=\int_0^lax\,dx=\frac{al^2}2 \); centre of mass at \( \frac{\int ax^2}{M}=\frac{2l}3 \).
2. \( I_{end}=\int_0^lax^3dx=\frac{al^4}4=\frac{Ml^2}2 \); \( I_{cm}=\frac{Ml^2}2-M\frac{4l^2}9=\frac{Ml^2}{18} \)

Teacher's Note:
1. Compute I about the end, then shift to the centre of mass.

Answer: (c) 3 m/s

Explanation:
1. \( I_1=\frac{0.75\times0.4^2}{12}+2(1)(0.1)^2=0.03,\quad I_2=0.01+2(1)(0.2)^2=0.09 \) (kg m squared).
2. Angular momentum: \( \omega_2=\frac{0.03\times30}{0.09}=10\ \text{rad/s} \).
3. Energy: \( \frac12(0.03)(900)=\frac12(0.09)(100)+2\cdot\frac12(1)v^2\Rightarrow13.5=4.5+v^2\Rightarrow v=3\ \text{m/s} \)

Teacher's Note:
1. Conserve angular momentum and energy; the energy not in rotation goes to radial motion.

Answer: (b) 210

Explanation:
1. At maximum elongation the velocity is perpendicular to the spring, so angular momentum gives \( 0.1\times11\times1=0.1\times v\times1.1\Rightarrow v=10\ \text{m/s} \).
2. Energy: \( \frac12(0.1)(121)=\frac12(0.1)(100)+\frac12k(0.1)^2 \) gives \( 1.05=0.005k\Rightarrow k=210\ \text{N/m} \)

Teacher's Note:
1. Conserve angular momentum and energy at the position of maximum elongation.

Answer: (b) \( \frac{2F}{M} \)

Explanation:
1. Centre of mass: \( a_{cm}=\frac FM \). Angular acceleration: \( \alpha=\frac{F\,L/2}{ML^2/12}=\frac{6F}{ML} \).
2. End B moves opposite to the force relative to the centre: \( a_B=\frac{F}{M}-\alpha\frac L2=\frac FM-\frac{3F}M=-\frac{2F}{M} \), magnitude 2F/M.

Teacher's Note:
1. Add the centre of mass acceleration and the rotational acceleration of the end.

Answer: (c) 1/4

Explanation:
1. Momentum: \( mv_0=MV \). Angular momentum about the centre: \( mv_0\frac L2=\frac{ML^2}{12}\omega\Rightarrow\omega=\frac{6mv_0}{ML} \).
2. Energy: \( \frac12mv_0^2=\frac12MV^2+\frac12\cdot\frac{ML^2}{12}\omega^2=\frac12\cdot\frac{4m^2v_0^2}{M} \), so \( M=4m \).

Teacher's Note:
1. Use momentum, angular momentum and energy together for an elastic collision.