JEE Physics System Of Particles and Rotational Motion MCQs Set 59

Answer: (d) \( \frac{300}{\pi} \)

Explanation:
1. \( K=\frac12I\omega^2 \) gives \( \omega=\sqrt{\frac{2\times1000}{20}}=10 \) rad/s.
2. Frequency \( =\frac{\omega}{2\pi}=\frac5\pi \) rev/s \( =\frac{300}{\pi} \) rev/min.

Teacher's Note:
1. Convert the angular velocity to revolutions per minute.

Answer: (d) \( 4MR^2 \)

Explanation:
1. Full disc: \( \frac12(9M)R^2=\frac92MR^2 \). Removed disc has mass \( M \) (area ratio \( \frac19 \)).
2. Removed part about the axis: \( \frac12M\left(\frac R3\right)^2+M\left(\frac{2R}{3}\right)^2=\frac{MR^2}{18}+\frac{4MR^2}{9}=\frac12MR^2 \).
3. \( I=\frac92MR^2-\frac12MR^2=4MR^2 \).

Teacher's Note:
1. Subtract the moment of inertia of the cut-out about the same axis.

Answer: (b) \( \frac37v_0 \) directed away from the wall

Explanation:
1. The wall is smooth, so the impact reverses only the centre velocity: \( v=-v_0 \) while the spin stays \( \omega=\frac{v_0}{R} \).
2. Angular momentum about a ground point is conserved afterwards: \( \frac25mR^2\frac{v_0}{R}-mv_0R=\frac75mv_fR \) (toward the wall positive).
3. \( v_f=-\frac37v_0 \), i.e. \( \frac37v_0 \) away from the wall.

Teacher's Note:
1. Friction passes through the ground contact, so angular momentum about it is conserved.

Answer: (b) only the direction of \( \vec L \) is constant

Explanation:
1. \( \vec L=m\vec r\times\vec v \) is perpendicular to the plane of the circle, always in the same direction.
2. Its magnitude \( mvr \) decreases because \( v \) decreases, and the acceleration has a tangential component, so it is not directed to the centre.

Teacher's Note:
1. Decreasing speed means a tangential acceleration and a changing magnitude of \( L \).

Answer: (c) remains unchanged

Explanation:
1. There is no external torque about the axis, so angular momentum is conserved.
2. Only the moment of inertia decreases (and \( \omega \) increases).

Teacher's Note:
1. Internal changes of the body do not change angular momentum.

Answer: (a) \( \frac{2g}{3} \)

Explanation:
1. \( Mg-T=Ma \) and \( TR=\frac12MR^2\cdot\frac aR \) give \( T=\frac{Ma}{2} \).
2. \( Mg=\frac32Ma \), so \( a=\frac{2g}{3} \).

Teacher's Note:
1. This is the yo-yo problem: the cylinder rolls down the string.

Answer: (c) \( \frac{r\omega_0}{2} \)

Explanation:
1. Angular momentum about a point on the ground is conserved: \( mr^2\omega_0=mvr+mr^2\frac vr \).
2. \( \omega_0r=2v \), so \( v=\frac{r\omega_0}{2} \).

Teacher's Note:
1. Friction has no torque about the contact point.

Answer: (a) \( 10 \) units

Explanation:
1. \( I_{diam}=\frac14MR^2=2 \), so \( MR^2=8 \).
2. About the tangent in the plane: \( I=\frac14MR^2+MR^2=\frac54\times8=10 \).

Teacher's Note:
1. The tangent in the plane of the disc is parallel to a diameter at distance \( R \).

Answer: (c) \( \frac{w(d-x)}{d} \)

Explanation:
1. Torque about B: \( N_Ad=w(d-x) \).
2. \( N_A=\frac{w(d-x)}{d} \) (and \( N_B=\frac{wx}{d} \)).

Teacher's Note:
1. Take torques about the other support.

Answer: (b) \( 10 \) J s

Explanation:
1. \( K=\frac{L^2}{2I} \), so \( L=\sqrt{2IK}=\sqrt{2\times5\times10}=10 \) J s.

Teacher's Note:
1. Use \( K=\frac{L^2}{2I} \).

Answer: (c) (i), (iii) and (iv)

Explanation:
1. Torque \( =mg\times \)(constant horizontal distance) is constant, so \( \frac{dL}{dt} \) is constant and \( L \) increases: (i) is true and (ii) is false.
2. The distance \( r \) of the particle from O decreases as it falls, so \( I=mr^2 \) decreases: (iii) true.
3. \( L=I\omega \) increases while \( I \) decreases, so \( \omega \) increases: (iv) true.

Teacher's Note:
1. Constant torque means \( L \) grows linearly in time.

Answer: (a) first increases and then decreases

Explanation:
1. No external torque: \( I\omega \) is constant.
2. As the insect moves towards the centre, \( I \) decreases so \( \omega \) increases; as it moves away to the other end, \( I \) increases so \( \omega \) decreases.

Teacher's Note:
1. Angular speed is inversely proportional to the moment of inertia.

Answer: (a) \( \frac{2g}{5} \)

Explanation:
1. \( a=\frac{mg}{m+m+\frac m2}=\frac{g}{2.5}=\frac{2g}{5} \).

Teacher's Note:
1. The pulley's inertia behaves as an extra mass \( \frac M2 \) in the string.

Answer: (b) the two sections of the string have unequal tensions

Explanation:
1. The string does not slip, so the pulley rotates with \( \alpha=\frac ar \); a net torque on a massive pulley needs \( T_1\neq T_2 \).
2. Both blocks share the same magnitude of acceleration \( a \).

Teacher's Note:
1. A massive pulley needs a tension difference to accelerate.

Answer: (b) \( \frac{v_0}{3} \) opposite to its initial direction

Explanation:
1. \( v_{cm}=\frac{mv_0}{3m}=\frac{v_0}{3} \). In the centre-of-mass frame A moves with \( v_0-\frac{v_0}{3}=\frac{2v_0}{3} \) and B with \( \frac{v_0}{3} \) the other way.
2. After \( 180^\circ \) the relative velocities reverse: A has \( v_{cm}-\frac{2v_0}{3}=-\frac{v_0}{3} \) (and B has \( \frac{2v_0}{3} \) in the initial direction).

Teacher's Note:
1. The centre moves uniformly while the pair rotates about it.

Answer: (a) \( 0.5 \) kg m\( ^2 \)/s

Explanation:
1. For a square plate about a diagonal, \( I=\frac{ma^2}{12}=\frac{2\times0.01}{12}=\frac{1}{600} \) kg m\( ^2 \).
2. \( L=\tau t=0.10\times5=0.5 \) kg m\( ^2 \)/s (and \( \omega=300 \) rad/s, kinetic energy \( 75 \) J).

Teacher's Note:
1. With constant torque, \( L=\tau t \) from rest.

Answer: (c) angular momentum

Explanation:
1. No external torque acts, so the angular momentum stays constant.
2. Changing the shape changes \( I \), and therefore \( \omega=\frac LI \) and \( K=\frac{L^2}{2I} \).

Teacher's Note:
1. Only external torques can change the angular momentum.

Answer: (d) \( -8 \)

Explanation:
1. \( \vec L=m(\vec r\times\vec v)=2\cdot2\,(\hat i+\hat j)\times(\hat i-\hat j+\hat k) \).
2. \( (\hat i+\hat j)\times(\hat i-\hat j+\hat k)=\hat i-\hat j-2\hat k \), so \( \vec L=4\hat i-4\hat j-8\hat k \).
3. The z-component is \( -8 \).

Teacher's Note:
1. Expand the cross product component by component.

Answer: (b) \( \sin\theta:2 \)

Explanation:
1. The string leaves the top of the rolling cylinder, whose top point moves at twice the centre speed, so B moves twice as far as A's centre along the incline.
2. A moves \( s \) along the incline (vertical component \( s\sin\theta \)) while B moves \( 2s \) vertically. Ratio \( =\sin\theta:2 \).

Teacher's Note:
1. The top of a rolling body moves with twice the speed of its centre.

Answer: (a) zero torque

Explanation:
1. A central force acts along the line joining the particle and the centre, so \( \vec r\times\vec F=0 \).
2. Zero torque means constant angular momentum.

Teacher's Note:
1. No torque means no change in angular momentum.

Answer: (c) the half lower down the incline

Explanation:
1. Friction acts at the base up the slope and tends to rotate the brick about its centre.
2. This torque is balanced by shifting the normal force towards the lower edge, so the lower half presses harder.

Teacher's Note:
1. The normal force acts closer to the lower edge.

Answer: (d) greater in process A than in process B

Explanation:
1. A: \( \alpha=\frac{FR}{I}=\frac{5(0.2)}{0.05}=20 \) rad/s\( ^2 \), so \( a=4 \) m/s\( ^2 \) (with \( I=\frac12MR^2=0.05 \)).
2. B: \( mg-T=ma \) and \( TR=I\frac aR \) give \( a=\frac{mg}{m+I/R^2}=\frac{5}{0.5+1.25}\approx2.86 \) m/s\( ^2 \).
3. So the acceleration in A is greater.

Teacher's Note:
1. The hanging body also has to be accelerated, which lowers the acceleration.

Answer: (b) \( 19.7 \) rad/s

Explanation:
1. \( I_{new}=0.5+0.2(0.2)^2=0.508 \) kg m\( ^2 \).
2. \( \omega'=\frac{I\omega}{I_{new}}=\frac{0.5\times20}{0.508}=19.7 \) rad/s.

Teacher's Note:
1. Angular momentum is conserved when the particle sticks to the rim.

Answer: (a) \( \frac{V}{2L} \)

Explanation:
1. Only the component of the impulse perpendicular to the rod, \( MV\sin30^\circ=\frac{MV}{2} \), produces rotation.
2. Angular momentum about the centre: \( \frac{MV}{2}\cdot\frac L2=I\omega \) with \( I=2M\left(\frac L2\right)^2=\frac{ML^2}{2} \).
3. \( \omega=\frac{V}{2L} \).

Teacher's Note:
1. Resolve the impulse into components along and perpendicular to the rod.

Answer: (d) \( \frac{ml^2}{12} \)

Explanation:
1. A rod about an axis through its centre at angle \( \theta \) to the rod: \( I=\frac{ml^2}{12}\sin^2\theta \).
2. Here \( \theta=45^\circ \), so each rod gives \( \frac{ml^2}{24} \) and the pair gives \( \frac{ml^2}{12} \).

Teacher's Note:
1. Only the component of the rod's length perpendicular to the axis contributes.

Answer: (d) \( \frac{Mv_0^2}{4} \)

Explanation:
1. Final rolling speed is \( \frac{v_0}{2} \), so \( K_f=\frac12Mv^2+\frac12MR^2\frac{v^2}{R^2}=Mv^2=\frac{Mv_0^2}{4} \).
2. \( K_i=\frac12Mv_0^2 \), so the loss is \( \frac{Mv_0^2}{4} \).

Teacher's Note:
1. Friction dissipates energy while the contact point slips.