JEE Physics System Of Particles and Rotational Motion MCQs Set 62

Answer: (a) \( \mu \gt \frac23 \)

Explanation:
1. Tipping about the edge: \( F\cdot\frac{3b}{4}=Mg\cdot\frac b2 \), so \( F=\frac{2Mg}{3} \).
2. Friction must supply \( F \) without slipping: \( \mu Mg \gt \frac{2Mg}{3} \), so \( \mu \gt \frac23 \).

Teacher's Note:
1. Compare the force needed to tip with the maximum friction.

Answer: (b) \( \frac{5500}{3} \) N

Explanation:
1. Friction on the rear wheel \( f=250\times2=500 \) N and \( N_r+N_f=2500 \) N.
2. Torque about the centre of mass: \( 500(1)+0.9N_f-0.6N_r=0 \).
3. \( 500+0.9(2500-N_r)-0.6N_r=0 \) gives \( N_r=\frac{2750}{1.5}=\frac{5500}{3} \) N.

Teacher's Note:
1. Take torques about the centre of mass, including the friction force.

Answer: (c) \( \frac{MV}{2} \)

Explanation:
1. Impulse \( J \): \( v_{cm}=\frac JM \) and \( \omega=\frac{J(L/2)}{ML^2/12}=\frac{6J}{ML} \).
2. Velocity of the far end \( =v_{cm}-\omega\frac L2=\frac JM-\frac{3J}{M}=-\frac{2J}{M} \), magnitude \( V=\frac{2J}{M} \), so \( J=\frac{MV}{2} \).

Teacher's Note:
1. Add the translational and rotational velocities of the far end.

Answer: (c) \( \sqrt2\,r \)

Explanation:
1. \( I=mr^2+mr^2=2mr^2 \) by the parallel axis theorem.
2. \( k=\sqrt{\frac Im}=\sqrt2r \).

Teacher's Note:
1. For a ring, \( I=2mr^2 \) about an axis through a point on the rim.

Answer: (b) \( \frac{10}{13} \) m/s\( ^2 \)

Explanation:
1. \( a=\frac{(m_A-m_B)g}{m_A+m_B+\frac M2}=\frac{2\times10}{4+2+20}=\frac{10}{13} \) m/s\( ^2 \).

Teacher's Note:
1. A massive disc pulley adds \( \frac M2 \) to the effective mass.

Answer: (b) \( 0.6 \)

Explanation:
1. Friction needed \( f=250\times6=1500 \) N and the normal reaction on the rear wheel is \( 2500 \) N.
2. \( \mu=\frac{f}{N}=\frac{1500}{2500}=0.6 \).

Teacher's Note:
1. The friction force must supply the full accelerating force.

Answer: (b) \( \frac hR \)

Explanation:
1. The string is inextensible and does not slip, so the length unwound equals the drop of the block, \( h \).
2. \( \Delta\theta=\frac{\text{arc}}{R}=\frac hR \).

Teacher's Note:
1. Angle equals unwound length divided by radius.

Answer: (c) \( \mu g\sqrt{\frac{3M}{k}} \)

Explanation:
1. Friction required: \( |f|=\frac{M|a|}{2}=\frac{2k|x|}{3} \), maximum at the amplitude \( A=\frac{V_0}{\omega} \) with \( \omega=\sqrt{\frac{4k}{3M}} \).
2. \( \frac{2k}{3}\cdot\frac{V_0}{\omega}\le\mu Mg \), so \( V_0\le\frac{3\mu Mg\omega}{2k}=\mu g\sqrt{\frac{3M}{k}} \).

Teacher's Note:
1. Rolling without slipping needs the required friction to stay below the limiting value.

Answer: (c) \( 1.58 \) m/s

Explanation:
1. Energy: \( mgx-\frac12kx^2=\frac12mv^2+\frac12\left(\frac12MR^2\right)\frac{v^2}{R^2} \).
2. \( 4(9.8)(0.2)-\frac12(80)(0.04)=7.84-1.6=6.24=\frac12(4)v^2+\frac14(2)v^2=2.5v^2 \).
3. \( v^2=2.496 \), so \( v\approx1.58 \) m/s.

Teacher's Note:
1. The pulley's rotational energy is \( \frac14Mv^2 \) for a disc.

Answer: (d) decrease

Explanation:
1. The mass is the same, but the mass is brought closer to the axis in the sphere.
2. Also the formula changes from \( \frac12MR^2 \) (large radius) to \( \frac25MR'^2 \) with smaller \( R' \). So \( I \) decreases.

Teacher's Note:
1. Compact shapes have smaller moments of inertia.

Answer: (a) \( l \)

Explanation:
1. Energy: \( \frac12\cdot\frac{ml^2}{3}\left(\frac{v_0}{l}\right)^2=\frac{mv_0^2}{6} \). This is less than \( mgl \) needed to reach the top (since \( v_0 \lt \sqrt{6gl} \)), so the rod swings up and stops.
2. The centre of mass rises \( \frac h2 \) when the end rises \( h \): \( mg\frac h2=\frac{mv_0^2}{6} \), so \( h=\frac{v_0^2}{3g}=l \).

Teacher's Note:
1. For \( v_0 \lt \sqrt{6gl} \) the rod does not complete a revolution; for larger \( v_0 \) the end rises \( 2l \).

Answer: (d) \( \frac{14}{5}mr^2 \)

Explanation:
1. For one sphere about a tangent: \( \frac25mr^2+mr^2=\frac75mr^2 \).
2. For two spheres: \( 2\times\frac75mr^2=\frac{14}{5}mr^2 \).

Teacher's Note:
1. Apply the parallel axis theorem to each sphere.

Answer: (a) \( 4 \) kg m\( ^2 \)

Explanation:
1. \( I=mr^2=4\times1^2=4 \) kg m\( ^2 \).
2. The rotation rate does not enter the moment of inertia.

Teacher's Note:
1. For a point mass \( I=mr^2 \).

Answer: (b) \( \frac{v}{2R} \)

Explanation:
1. Angular momentum about O: \( mvR=I\omega \) with \( I=mR^2+mR^2=2mR^2 \) (ring and bead are both at distance \( R \)).
2. \( \omega=\frac{v}{2R} \).

Teacher's Note:
1. A semicircular ring about its centre has \( I=MR^2 \), like a full ring.

Answer: (a) \( \frac{v_1l}{v_1+v_2} \)

Explanation:
1. \( \omega=\frac{v_1+v_2}{l} \) (opposite directions).
2. The point at rest is at distance \( \frac{v_1}{\omega}=\frac{v_1l}{v_1+v_2} \) from the end of speed \( v_1 \).

Teacher's Note:
1. The instantaneous axis lies where the velocity is zero, between the two ends.

Answer: (c) \( \frac{g\sin\theta}{1+\frac{I}{MR^2}} \)

Explanation:
1. \( Mg\sin\theta-f=Ma \) and \( fR=I\frac aR \).
2. \( a=\frac{g\sin\theta}{1+\frac{I}{MR^2}} \).

Teacher's Note:
1. Rolling friction supplies the torque for the angular acceleration.

Answer: (d) \( \frac{50}{9\pi} \) and \( \frac{20}{3} \) m

Explanation:
1. \( \omega_0=\frac{2}{0.9} \) rad/s and \( \alpha=\frac{3}{0.9}=\frac{10}{3} \) rad/s\( ^2 \).
2. \( \theta=\omega_0t+\frac12\alpha t^2=\frac{4}{0.9}+\frac12\cdot\frac{10}{3}\cdot4=\frac{40}{9}+\frac{20}{3}=\frac{100}{9} \) rad.
3. Revolutions \( =\frac{100}{9\cdot2\pi}=\frac{50}{9\pi} \).
4. Load A moves on the radius \( 0.6 \) m: \( s=0.6\times\frac{100}{9}=\frac{20}{3} \) m.

Teacher's Note:
1. Convert the linear motion of B into angular motion of the pulley.

Answer: (c) \( \sqrt{\frac{3g}{l}} \)

Explanation:
1. Centre of mass rises \( \frac l2 \): \( \frac12\cdot\frac{ml^2}{3}\omega^2=mg\frac l2 \).
2. \( \omega=\sqrt{\frac{3g}{l}} \).

Teacher's Note:
1. Rotational kinetic energy equals the gain in potential energy.

Answer: (b) \( \frac13 \) N m

Explanation:
1. \( \vec a=t\hat i+\hat j \), so \( \vec v=\frac{t^2}{2}\hat i+t\hat j \) and \( \vec r=\frac{t^3}{6}\hat i+\frac{t^2}{2}\hat j \).
2. At \( t=1 \): \( \tau_z=xF_y-yF_x=\frac16-\frac12=-\frac13 \), magnitude \( \frac13 \) N m (direction \( -\hat k \)).

Teacher's Note:
1. Integrate twice, then take \( \vec r\times\vec F \).

Answer: (b) greater than \( \frac I2 \)

Explanation:
1. With \( PQ=a \), \( QR=b \) and mass \( m \) for the whole plate: \( I=\frac{m(a^2+b^2)}{3} \).
2. Triangle PQR (mass \( \frac m2 \)) has \( I_P=\frac m2\left(\frac{a^2}{2}+\frac{b^2}{6}\right) \), which exceeds \( \frac I2=\frac{m}{2}\cdot\frac{a^2+b^2}{3} \) when \( a \gt b \).
3. Hence it is greater than \( \frac I2 \).

Teacher's Note:
1. The half containing the far long edge has the larger moment about P.

Answer: (d) \( 4R \)

Explanation:
1. At the highest point the particle moves with speed \( 2v \) and its acceleration is \( \frac{v^2}{R} \) (towards the centre, perpendicular to velocity).
2. \( \rho=\frac{(2v)^2}{v^2/R}=4R \).

Teacher's Note:
1. The cycloid has radius of curvature \( 4R \) at its crest.

Answer: (d) \( \frac{ML^2}{6} \)

Explanation:
1. The height is \( \frac L2 \). About the x-axis (through the apex, parallel to the base): \( I_x=\frac{M}{2}\left(\frac L2\right)^2=\frac{ML^2}{8} \).
2. About the y-axis (axis of symmetry): \( I_y=\frac{ML^2}{24} \).
3. \( I_z=I_x+I_y=\frac{ML^2}{8}+\frac{ML^2}{24}=\frac{ML^2}{6} \).

Teacher's Note:
1. Use the perpendicular axis theorem.

Answer: (a) \( -\frac{20}{3}\hat k \) N m

Explanation:
1. \( \vec a=20\hat i+10\hat j \), so \( \vec F=m\vec a=2\hat i+1\hat j \) N.
2. \( \tau_z=xF_y-yF_x=\frac{10}{3}(1)-5(2)=-\frac{20}{3} \).

Teacher's Note:
1. Torque is \( \vec r\times\vec F \).

Answer: (a) all take the same time

Explanation:
1. With no friction there is no torque, so none of them rotates.
2. Each slides with acceleration \( g\sin\theta \), independent of shape and mass, so the times are equal.

Teacher's Note:
1. Smooth surfaces mean pure sliding, not rolling.

Answer: (a) \( \frac{3A_0}{4} \)

Explanation:
1. \( \tau=\frac{\Delta L}{\Delta t}=\frac{4A_0-A_0}{4}=\frac{3A_0}{4} \).

Teacher's Note:
1. Torque is the rate of change of angular momentum.

Answer: (d) \( 6 \) m/s\( ^2 \)

Explanation:
1. The front wheel just lifts when \( N_f=0 \), so \( N_r=2500 \) N.
2. Torque about the centre of mass: \( 250a(1)=N_r(0.6)=1500 \), so \( a=6 \) m/s\( ^2 \).

Teacher's Note:
1. The front wheel leaves the ground when the normal force on it becomes zero.