JEE Physics System Of Particles and Rotational Motion MCQs Set 61

Answer: (a) \( \frac{4MR^2}{9\sqrt3\,\pi} \)

Explanation:
1. The body diagonal of the cube is \( 2R \), so the edge is \( a=\frac{2R}{\sqrt3} \).
2. Mass of the cube \( =M\frac{a^3}{\frac43\pi R^3}=\frac{2M}{\sqrt3\pi} \).
3. \( I=\frac{M_ca^2}{6}=\frac{2M}{\sqrt3\pi}\cdot\frac{4R^2}{3\cdot6}=\frac{4MR^2}{9\sqrt3\pi} \).

Teacher's Note:
1. The mass of the cube is its volume fraction times \( M \).

Answer: (c) \( \frac{(I+mR^2)\omega-mvR}{I} \)

Explanation:
1. Angular momentum is conserved: \( (I+mR^2)\omega=I\omega'+mvR \).
2. \( \omega'=\frac{(I+mR^2)\omega-mvR}{I} \).

Teacher's Note:
1. The child's contribution to the angular momentum initially is \( mR^2\omega \).

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

Explanation:
1. The solid cylinder has acceleration \( \frac23g\sin\theta \) and the hollow one \( \frac12g\sin\theta \), so the solid cylinder reaches the bottom first: A is false.
2. Both lose the same potential energy \( mgh \), so their total kinetic energies are equal: R is true.

Teacher's Note:
1. A smaller moment of inertia means a larger share of energy goes into translation, so the solid cylinder is faster.

Answer: (d) less than \( 1 \)

Explanation:
1. \( I_1=\frac{ML^2}{12} \).
2. The arc has radius \( R=\frac L\pi \), so \( I_2=MR^2=\frac{ML^2}{\pi^2} \).
3. \( \frac{I_1}{I_2}=\frac{\pi^2}{12}\approx0.82 \lt 1 \).

Teacher's Note:
1. For a bent rod all the mass is at distance \( R \) from the axis.

Answer: (d) \( -\frac{4kx}{3} \)

Explanation:
1. The springs give \( -2kx \), and friction \( f \) gives the torque for rolling: \( fR=\frac12MR^2\frac aR \) in the opposite sense, so \( f=-\frac{Ma}{2} \).
2. \( Ma=-2kx-\frac{Ma}{2} \), so \( Ma=-\frac{4kx}{3} \) (the net external force).

Teacher's Note:
1. For a rolling disc, the friction contributes \( -\frac{Ma}{2} \) to the force equation.

Answer: (c) \( \frac13mg\sin\theta \) up the incline

Explanation:
1. \( a=\frac{2}{3}g\sin\theta \) for a solid cylinder.
2. \( mg\sin\theta-f=ma \) gives \( f=\frac13mg\sin\theta \), directed up the incline.

Teacher's Note:
1. The friction is static, so it is not \( \mu N \); it is whatever rolling requires.

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

Explanation:
1. \( a_{cm}=\frac FM \) and \( \alpha=\frac{F\frac L2}{ML^2/12}=\frac{6F}{ML} \).
2. End B: \( a_B=a_{cm}-\alpha\frac L2=\frac FM-\frac{3F}{M}=-\frac{2F}{M} \), so the magnitude is \( \frac{2F}{M} \) (opposite to the force).

Teacher's Note:
1. The far end moves backwards because the rotational part exceeds the translational part.

Answer: (c) a circular arc of radius \( \frac L2 \) about the corner

Explanation:
1. While both ends touch the wall and floor, the midpoint is the midpoint of the hypotenuse of a right triangle with the corner as the right-angle vertex, so it stays at distance \( \frac L2 \) from the corner.
2. So the path is a circle of radius \( \frac L2 \).

Teacher's Note:
1. The median to the hypotenuse equals half the hypotenuse.

Answer: (a) remains constant

Explanation:
1. \( L=mvy \) where \( y \) is the fixed perpendicular distance of the line of motion from the origin.
2. Both \( m \), \( v \) and \( y \) are constant, so \( L \) is constant.

Teacher's Note:
1. \( L=mv\times \) (perpendicular distance) is constant for straight-line motion at constant speed.

Answer: (b) \( \frac12Mga\sin\theta \)

Explanation:
1. Uniform velocity means no net force and no net torque. Friction \( f=Mg\sin\theta \) acts at the base, at distance \( \frac a2 \) below the centre, so its torque is \( \frac12Mga\sin\theta \).
2. The normal force must supply an equal and opposite torque: \( \frac12Mga\sin\theta \).

Teacher's Note:
1. The normal force shifts its line of action to balance the torque of friction.

Answer: (a) \( \frac{2I\omega}{3t} \)

Explanation:
1. Conservation of angular momentum: \( I(2\omega)+2I\omega=3I\omega' \), so \( \omega'=\frac{4\omega}{3} \).
2. For disc A: \( \Delta L=I\left(2\omega-\frac{4\omega}{3}\right)=\frac{2I\omega}{3} \), so \( \tau=\frac{2I\omega}{3t} \).

Teacher's Note:
1. Torque is the change of angular momentum of one disc divided by the time.

Answer: (b) \( \frac13 \)

Explanation:
1. Translational \( K=\frac12mv^2 \), rotational \( K=\frac12\cdot\frac12mR^2\cdot\frac{v^2}{R^2}=\frac14mv^2 \).
2. Ratio \( =\frac{1/4}{3/4}=\frac13 \).

Teacher's Note:
1. For a rolling disc, \( K_{rot}:K_{trans}=1:2 \).

Answer: (b) \( -14\hat i+10\hat j-9\hat k \)

Explanation:
1. \( \vec r_1=(2,1,-1) \) and \( \vec r_2=(1,1,1) \) relative to the point.
2. \( \vec\tau_1=\vec r_1\times\vec F_1=-11\hat i+10\hat j-12\hat k \) and \( \vec\tau_2=\vec r_2\times\vec F_2=-3\hat i+3\hat k \).
3. Total \( =-14\hat i+10\hat j-9\hat k \).

Teacher's Note:
1. Take position vectors relative to the chosen point and add the two cross products.

Answer: (d) \( w\ell\sin\theta \)

Explanation:
1. \( \tau=r_\perp F=(\ell\sin\theta)w \).
2. The torque is zero when \( \theta=0 \), i.e. when the bob is at its lowest point.

Teacher's Note:
1. The lever arm of the weight is the horizontal distance \( \ell\sin\theta \).

Answer: (a) \( \sqrt{\frac{3g}{5L}} \)

Explanation:
1. The centre of mass falls \( \frac L2(1-\cos37^\circ)=\frac L{10} \).
2. \( \frac12\cdot\frac{ML^2}{3}\omega^2=Mg\frac{L}{10} \), so \( \omega^2=\frac{3g}{5L} \).

Teacher's Note:
1. Apply energy conservation with the centre-of-mass drop.

Answer: (c) \( \frac{mg}{2}\cot\theta \)

Explanation:
1. Vertical forces: \( N_{floor}=mg \). Horizontal forces: \( P=N_{wall} \).
2. Torque about the lower end: \( N_{wall}L\sin\theta=mg\frac L2\cos\theta \), so \( N_{wall}=\frac{mg}{2}\cot\theta=P \).

Teacher's Note:
1. Take torques about the foot of the rod to eliminate the floor reaction.

Answer: (b) \( \frac{181}{2}MR^2 \)

Explanation:
1. About O: central disc \( \frac12MR^2 \); each outer disc \( \frac12MR^2+M(2R)^2=\frac92MR^2 \), so six give \( 27MR^2 \). Total \( =\frac{55}{2}MR^2 \).
2. Parallel axis to P (\( OP=3R \), total mass \( 7M \)): \( I_P=\frac{55}{2}MR^2+7M(3R)^2=\frac{55}{2}MR^2+63MR^2=\frac{181}{2}MR^2 \).

Teacher's Note:
1. First find \( I \) about the centre of the arrangement, then shift the axis.

Answer: (b) \( \frac L4 \)

Explanation:
1. The centre of mass falls vertically. With angle \( \phi \) to the floor, the point at distance \( s \) from the lower end has \( x=\left(\frac L2-s\right)\cos\phi \) and \( y=s\sin\phi \).
2. This is a circle when \( \left|\frac L2-s\right|=s \), i.e. \( s=\frac L4 \).

Teacher's Note:
1. A point traces a circle only if the semi-axes of its elliptical path are equal.

Answer: (d) \( \frac{24}{289} \) rad/s

Explanation:
1. After \( 5 \) s the position is \( (15,8) \) and \( r^2=15^2+8^2=289 \).
2. \( \omega=\frac{|xv_y-yv_x|}{r^2}=\frac{8\times3}{289}=\frac{24}{289} \) rad/s.

Teacher's Note:
1. \( \omega=\frac{v_\perp}{r}=\frac{vy}{r^2} \) for straight-line motion.

Answer: (a) \( \frac34 \)

Explanation:
1. Angular momentum about the hinge: \( mv\frac L2=\frac{ML^2}{3}\omega \), so \( \omega=\frac{3mv}{2ML} \).
2. Elastic: \( \frac12mv^2=\frac12\cdot\frac{ML^2}{3}\omega^2=\frac{3m^2v^2}{8M} \).
3. \( 1=\frac{3m}{4M} \), so \( \frac Mm=\frac34 \).

Teacher's Note:
1. Use both angular momentum conservation about the hinge and energy conservation.

Answer: (c) \( 10 \) m/s

Explanation:
1. Area of the struck half \( =\frac{ab}{2}=1 \) m\( ^2 \). Each ball transfers impulse \( 2mv \), so force \( F=n\cdot1\cdot2mv=2v \) N, acting at the centre of that half, \( \frac{3b}{4}=1.5 \) m from the hinge.
2. Torque balance about the hinge: \( F(1.5)=Mg\frac b2=30 \) N m, so \( F=20 \) N.
3. \( 2v=20 \) gives \( v=10 \) m/s.

Teacher's Note:
1. Take torques about the hinge edge: weight at the centre, ball force at the centre of the struck half.

Answer: (a) \( \frac{v_B-v_A}{d} \)

Explanation:
1. Both velocities are proportional to the distance from the instantaneous centre: \( v_A=\omega r_A \) and \( v_B=\omega r_B \) with \( r_B-r_A=d \).
2. \( \omega=\frac{v_B-v_A}{d} \).

Teacher's Note:
1. Subtract the two velocity relations to eliminate the unknown distance.

Answer: (b) \( \frac{MR^2}{2}-M\left(\frac{4R}{3\pi}\right)^2 \)

Explanation:
1. About the centre of the full circle the half-disc has \( I=\frac12MR^2 \) (same as a full disc of the same density per unit mass).
2. The centre of mass is \( \frac{4R}{3\pi} \) from that centre. Parallel axis theorem: \( I_{cm}=\frac12MR^2-M\left(\frac{4R}{3\pi}\right)^2 \).

Teacher's Note:
1. Move to the centre of mass by subtracting \( Md^2 \).

Answer: (b) more than \( 3 \) but less than \( 6 \)

Explanation:
1. \( \alpha=\frac{2(20t-5t^2)}{10}=4t-t^2 \), so \( \omega=2t^2-\frac{t^3}{3} \).
2. \( \omega=0 \) at \( t=6 \) s, when the direction reverses. Angle \( =\int_0^6\omega\,dt=\left[\frac{2t^3}{3}-\frac{t^4}{12}\right]_0^6=144-108=36 \) rad.
3. Rotations \( =\frac{36}{2\pi}\approx5.7 \).

Teacher's Note:
1. Integrate the angular acceleration twice up to the time when \( \omega=0 \).

Answer: (d) \( \frac12KE_0 \)

Explanation:
1. Angular momentum: \( I\omega_0=2I\omega' \), so \( \omega'=\frac{\omega_0}{2} \).
2. \( K'=\frac12(2I)\left(\frac{\omega_0}{2}\right)^2=\frac14I\omega_0^2=\frac12KE_0 \).

Teacher's Note:
1. Half of the energy is lost in the inelastic 'rotational' collision.

Answer: (a) \( -\frac53\hat k \) N m s

Explanation:
1. \( \vec v=3\alpha t^2\hat i+2\beta t\hat j=10\hat i+10\hat j \) and \( \vec r=\frac{10}{3}\hat i+5\hat j \) at \( t=1 \).
2. \( L_z=m(xv_y-yv_x)=0.1\left(\frac{100}{3}-50\right)=-\frac53 \).

Teacher's Note:
1. Use \( \vec L=m\,\vec r\times\vec v \).