JEE Physics System Of Particles and Rotational Motion MCQs Set 47

Answer: (b) \( \dfrac{3g}{4L} \)

Explanation:
1. Rod pivots about A (no sliding). Torque of the weight: \( mg\dfrac L2\cos\beta \). Moment of inertia about A: \( \dfrac{mL^2}{3} \).
2. \( \alpha=\dfrac{3g\cos\beta}{2L}=\dfrac{3g}{2L}\times\dfrac12=\dfrac{3g}{4L} \).

Teacher's Note:
1. Initially the rod is at rest, so only the tangential acceleration exists.
2. Use the lever arm L/2 times cos beta for the weight.

Answer: (a) \( \dfrac{13}{16}mg \) and \( \dfrac{3\sqrt3}{16}mg \)

Explanation:
1. The centre of mass moves on a circle about A with tangential acceleration \( a=\alpha\dfrac L2=\dfrac{3g}{8} \), perpendicular to the rod.
2. Vertical: \( mg-N=ma\cos\beta=\dfrac{3mg}{16}\Rightarrow N=\dfrac{13}{16}mg \).
3. Horizontal: \( f=ma\sin\beta=\dfrac{3mg}{8}\times\dfrac{\sqrt3}{2}=\dfrac{3\sqrt3}{16}mg \).

Teacher's Note:
1. Resolve the tangential acceleration of the centre of mass into vertical and horizontal parts.
2. The centripetal term is zero because the angular velocity is zero.

Answer: (c) \( F\left(\dfrac{3x}{2L}-1\right) \)

Explanation:
1. \( \alpha=\dfrac{Fx}{mL^2/3}=\dfrac{3Fx}{mL^2} \), so the centre of mass accelerates (perpendicular to the rod) with \( a=\alpha\dfrac L2=\dfrac{3Fx}{2mL} \).
2. Along F: \( F+N=ma \Rightarrow N=\dfrac{3Fx}{2L}-F=F\left(\dfrac{3x}{2L}-1\right) \). It is zero at \( x=\dfrac{2L}{3} \) (centre of percussion).

Teacher's Note:
1. The angular velocity is zero, so there is no centripetal term.
2. Apply Newton's second law to the centre of mass along the force.

Answer: (c) \( \dfrac32\sqrt{\dfrac gL} \)

Explanation:
1. Moment of inertia about the leg: leg 0, crossbar (end on axis) \( \dfrac{mL^2}{3} \), far leg \( mL^2 \). \( I=\dfrac{4mL^2}{3} \).
2. Centre of mass falls: crossbar \( \dfrac L2 \), far leg \( L \). \( \Delta U=mg\dfrac L2+mgL=\dfrac32mgL \).
3. \( \dfrac12I\omega^2=\dfrac32mgL\Rightarrow\omega^2=\dfrac{9g}{4L} \), \( \omega=\dfrac32\sqrt{g/L} \).

Teacher's Note:
1. The leg on the axis contributes nothing to the moment of inertia.
2. Add the drops of the centres of mass of the crossbar and of the far leg.

Answer: (b) \( \dfrac12m\omega^2\left(R^2+\dfrac{L^2}{12}\right) \)

Explanation:
1. The rod rotates rigidly about the disc axis with \( \omega \). Its moment of inertia about that axis is \( \dfrac{mL^2}{12}+mR^2 \) (parallel axes; the rod is along the radius, so the \( \dfrac{mL^2}{12} \) term is about its centre).
2. \( K=\dfrac12I\omega^2=\dfrac12m\omega^2\left(R^2+\dfrac{L^2}{12}\right) \).

Teacher's Note:
1. Treat it as a rigid body rotating about the disc axis.
2. Check the limiting case R = 0 or L small.

Answer: (a) 2 m/s

Explanation:
1. Energy: \( mgh=\dfrac12mv^2+\dfrac12I\left(\dfrac vR\right)^2 \).
2. \( 3\times10\times0.4=\dfrac12(3)v^2+\dfrac12(3)v^2=3v^2 \Rightarrow v^2=4 \), \( v=2 \) m/s.

Teacher's Note:
1. The pulley turns at v over R.
2. Gravitational energy lost goes to both the block and the pulley.

Answer: (d) \( 2\hat k \) kg m²/s

Explanation:
1. \( \vec p=m\vec v=4\hat i+6\hat j \) and \( \vec r=\hat i+\hat j \).
2. \( \vec L=\vec r\times\vec p=(1\times6-1\times4)\hat k=2\hat k \) kg m²/s.

Teacher's Note:
1. Use L = r cross p.
2. Both vectors lie in the xy-plane, so L is along z.

Answer: (d) 16 kg m²/s

Explanation:
1. Perpendicular distance of the line from the origin: \( d=\dfrac{5}{\sqrt{3^2+4^2}}=1 \) m.
2. \( L=mvd=2\times8\times1=16 \) kg m²/s.

Teacher's Note:
1. Angular momentum of a particle on a straight path is m v times the perpendicular distance of the path from the point.
2. The distance of a line ax+by=c from the origin is c over root of a squared plus b squared.

Answer: (b) 10 rad/s

Explanation:
1. \( I=2\times2\times(0.5)^2=1 \) kg m². Initial \( L=I\omega=9 \) kg m²/s.
2. Angular impulse \( =F\,r\,t=10\times0.5\times0.2=1 \) kg m²/s.
3. \( \omega'=\dfrac{9+1}{1}=10 \) rad/s.

Teacher's Note:
1. The force is perpendicular to the radius.
2. Angular impulse equals the change in angular momentum.

Answer: (a) \( \dfrac{4\pi m}{M+2m} \)

Explanation:
1. Let the board turn at \( \omega \) opposite to the man. Man's speed relative to the ground is \( v_0-\omega R \). Angular momentum conservation: \( mR(v_0-\omega R)=\dfrac12MR^2\omega \), so \( \omega=\dfrac{2mv_0}{R(M+2m)} \).
2. Time to pass his starting point on the board: relative angle \( 2\pi \) at rate \( v_0/R \): \( t=\dfrac{2\pi R}{v_0} \).
3. Board's angle \( =\omega t=\dfrac{4\pi m}{M+2m} \).

Teacher's Note:
1. The man's speed relative to the ground is smaller than v0.
2. Use the relative angular speed to find the time.

Answer: (d) \( v_A=75 \) m/s, \( v_B=25 \) m/s

Explanation:
1. Speed due to rotation of each end relative to the centre: \( \omega\dfrac L2=5\times5=25 \) m/s.
2. Clockwise rotation moves the top end A to the right (same as the translation) and the bottom end B to the left.
3. \( v_A=50+25=75 \) m/s, \( v_B=50-25=25 \) m/s.

Teacher's Note:
1. Velocity of a point = velocity of the centre plus omega cross r.
2. The two ends have opposite rotational velocities.

Answer: (c) \( -2\hat i \) m/s

Explanation:
1. Anticlockwise rotation gives the lowest point a velocity \( +\omega R=+3 \) m/s along x relative to the centre.
2. \( v_P=v_{cm}+3\Rightarrow 1=v_{cm}+3\Rightarrow v_{cm}=-2 \) m/s.

Teacher's Note:
1. Velocity of lowest point = v of centre + omega cross r with r pointing down.
2. For anticlockwise rotation the bottom moves to the right relative to the centre.

Answer: (b) 7 m/s

Explanation:
1. Clockwise rotation gives the bottom point a velocity \( \omega R=3 \) m/s to the left relative to the centre.
2. No slipping: bottom point moves with the plank: \( v_{cm}-3=4\Rightarrow v_{cm}=7 \) m/s.

Teacher's Note:
1. The contact point of the disc has the same velocity as the plank.
2. Clockwise rotation moves the lowest point backward relative to the centre.

Answer: (c) \( \dfrac{5v_0}{6} \)

Explanation:
1. The centre has velocity components \( \dfrac{v_0}{2} \) horizontally and \( \dfrac{v_A}{2}=\dfrac{2v_0}{3} \) vertically.
2. \( v_{cm}=v_0\sqrt{\dfrac14+\dfrac49}=\dfrac{5v_0}{6} \). (Also \( \omega\,\dfrac{\ell}{2}=\dfrac{5v_0}{6} \).)

Teacher's Note:
1. The centre of mass is the midpoint of A and B.
2. Average the velocities of A and B.

Answer: (b) 5.59 cm/s², 2.50 cm/s², 2.50 cm/s²

Explanation:
1. \( \alpha=a/R=0.25 \) rad/s² and centripetal acceleration \( \omega^2R=2.50 \) cm/s² toward the centre.
2. A: forward \( a+\alpha R=5.00 \) and downward 2.50, so \( \sqrt{25+6.25}=5.59 \). B: forward \( 2.50-2.50=0 \) and downward \( \alpha R=2.50 \), so 2.50. O: only the centripetal 2.50 (upward).

Teacher's Note:
1. Add the acceleration of the centre, the tangential part alpha R and the centripetal part omega squared R.
2. The point of contact accelerates, though its velocity is zero.

Answer: (a) \( \sqrt{\dfrac{4gh}{3}} \)

Explanation:
1. Energy conservation (rolling, no loss): \( mgh=\dfrac12mv^2+\dfrac12\left(\dfrac12mR^2\right)\dfrac{v^2}{R^2}=\dfrac34mv^2 \).
2. \( v=\sqrt{4gh/3} \).

Teacher's Note:
1. A solid cylinder has I = (1/2) m R squared.
2. Rolling without slipping ties v to omega R.

Answer: (d) \( \dfrac{7}{10}mV^2 \)

Explanation:
1. Speed of lowest point relative to the centre is \( \omega R=V \), and rolling without sliding gives \( v_{cm}=V \).
2. \( K=\dfrac12mV^2+\dfrac12\cdot\dfrac25mR^2\cdot\dfrac{V^2}{R^2}=\dfrac{7}{10}mV^2 \).

Teacher's Note:
1. Add translational and rotational kinetic energies.
2. Solid sphere: I = (2/5) m R squared.

Answer: (b) \( \sqrt{3g(R-r)} \)

Explanation:
1. At the top, the minimum condition for the centre (moving on radius \( R-r \)) is \( mg=\dfrac{mv_t^2}{R-r} \), so \( v_t^2=g(R-r) \).
2. Energy for the ring (\( K=mv^2 \)): \( mv_0^2=mv_t^2+mg\cdot2(R-r) \).
3. \( v_0^2=3g(R-r) \).

Teacher's Note:
1. A rolling ring has translational and rotational energy equal.
2. The centre of the ring rises 2(R - r).

Answer: (a) \( \dfrac{4\pi R}{5} \)

Explanation:
1. Surface is smooth, so only F acts horizontally: \( a=\dfrac Fm \) and \( \alpha=\dfrac{FR}{\frac25mR^2}=\dfrac{5F}{2mR} \).
2. One revolution: \( 2\pi=\dfrac12\alpha t^2\Rightarrow t^2=\dfrac{8\pi mR}{5F} \).
3. \( s=\dfrac12at^2=\dfrac12\cdot\dfrac Fm\cdot\dfrac{8\pi mR}{5F}=\dfrac{4\pi R}{5} \).

Teacher's Note:
1. No friction, so translation and rotation are independent.
2. Use the same time for both motions.

Answer: (a) 20 N

Explanation:
1. With \( I=\dfrac23mR^2 \), rolling: \( F+f=\dfrac23ma \) and \( F-f=ma \) (f taken backward). Adding: \( a=\dfrac{6F}{5m} \), so \( f=F-ma=-\dfrac F5 \): friction acts forward with magnitude \( F/5 \).
2. Require \( \dfrac F5\le\mu mg=0.4\times1\times10=4 \) N, so \( F_{max}=20 \) N.

Teacher's Note:
1. Write the translational and rotational equations with rolling condition a = alpha R.
2. The friction needed is a fraction of F.

Answer: (d) \( \dfrac{3v}{2\ell} \)

Explanation:
1. Linear momentum: \( mv=4mV\Rightarrow V=\dfrac v4 \).
2. Angular momentum about the centre of the rod: \( mv\dfrac\ell2=\dfrac{4m\ell^2}{12}\omega \).
3. \( \omega=\dfrac{3v}{2\ell} \).

Teacher's Note:
1. Both linear and angular momentum are conserved.
2. Take angular momentum about the centre of the rod.

Answer: (c) \( \dfrac{m_2u}{m_1+m_2} \)

Explanation:
1. No external horizontal force, so momentum is conserved: \( m_2u=(m_1+m_2)v_C \).
2. \( v_C=\dfrac{m_2u}{m_1+m_2} \).

Teacher's Note:
1. The velocity of the centre of mass does not change in the collision except through momentum conservation.
2. Total momentum is that of the particle alone.

Answer: (b) \( \dfrac{m_1u}{m_1+m_2} \) and \( \dfrac{m_2u}{m_1+m_2} \)

Explanation:
1. \( v_C=\dfrac{m_2u}{m_1+m_2} \) (the rod is at rest, the particle moves at u).
2. Particle relative to C: \( u-v_C=\dfrac{m_1u}{m_1+m_2} \). Rod relative to C: \( 0-v_C \), speed \( \dfrac{m_2u}{m_1+m_2} \).

Teacher's Note:
1. Subtract the velocity of the centre of mass from each velocity.
2. The rod is at rest before the collision.