JEE Physics System Of Particles and Rotational Motion MCQs Set 68

Answer: (c) Both reach the wall at the same time

Explanation:
1. Total KE \( =\tfrac12mv^2(1+k) \). Ring: \( K=mv^2 \Rightarrow v^2=\dfrac K{0.3} \).
2. Cylinder: \( K=\tfrac34mv^2\Rightarrow v^2=\dfrac K{0.75\times0.4}=\dfrac K{0.3} \). The speeds are equal, so the times are equal.

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Answer: (d) 7 m/s

Explanation:
1. Rolling disc: \( KE=\tfrac34mv^2 \). Energy: \( \tfrac34v_1^2+gh_1=\tfrac34v_2^2+gh_2 \) (same final speed).
2. \( \tfrac34(9)+300=\tfrac34v_2^2+270\Rightarrow v_2^2=49\Rightarrow v_2=7 m/s \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Answer: (a) 3.6

Explanation:
1. Let \( N \) be the horizontal normal force from the stick and \( F_s \) its vertical friction, with \( N^2+F_s^2=4 \).
2. Linear: \( N-f_g=ma=0.6 \). Rotation about the centre (rolling, \( \alpha=a/R=0.6 \)): \( R(f_g-F_s)=mR^2\alpha\Rightarrow f_g-F_s=0.6 \), so \( F_s=N-1.2 \).
3. \( N^2+(N-1.2)^2=4\Rightarrow N\approx1.88 N \), \( F_s\approx0.68 N \). \( \mu=\dfrac{0.68}{1.88}=0.36 \), so \( P = 3.6 \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Answer: (b) \( \dfrac23g\sin\theta \)

Explanation:
1. \( Mg\sin\theta-f=Ma \), \( fR=\tfrac12MR^2\dfrac aR\Rightarrow f=\tfrac12Ma \).
2. \( a=\dfrac{g\sin\theta}{1+\tfrac12}=\dfrac23g\sin\theta \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Answer: (c) \( \dfrac{8F}{3m_1+8m_2} \)

Explanation:
1. The top point of a rolling cylinder moves at twice the centre speed, so \( a_{plank}=2a_{cm} \).
2. Cylinder (friction \( f \) from plank on top, \( f_g \) from floor): \( m_1a=f+f_g \) and \( \tfrac12m_1a=f-f_g \), giving \( f=\tfrac34m_1a \).
3. Plank: \( F-f=m_2(2a)\Rightarrow a=\dfrac{4F}{3m_1+8m_2} \), so \( a_{plank}=\dfrac{8F}{3m_1+8m_2} \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Answer: (d) \( \dfrac{3Fm_1}{3m_1+8m_2} \)

Explanation:
1. From the previous result \( a_{cm}=\dfrac{4F}{3m_1+8m_2} \) and \( f=\tfrac34m_1a_{cm} \).
2. \( f=\dfrac{3m_1F}{3m_1+8m_2} \). (The floor friction is \( \dfrac{Fm_1}{3m_1+8m_2} \).)

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Answer: (a) 2.14 m

Explanation:
1. Rolling from a drop of \( 1.6 m \): \( mg(1.6)=\tfrac12mv^2\left(1+\tfrac25\right)\Rightarrow v^2=\dfrac{10(1.6)}{0.7}=22.86 \), \( v=4.78 m/s \).
2. Fall time from \( 1.0 m \): \( t=\sqrt{\dfrac{2(1.0)}{10}}=0.447 s \).
3. Range \( =vt=4.78\times0.447\approx2.14 m \). In flight no torque acts, so the sphere keeps rotating about its centre of mass at constant angular speed.

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Answer: (b) \( \dfrac23v_0 \)

Explanation:
1. Angular momentum about a point on the floor is conserved (friction acts at that point): \( mv_0R=mvR+\tfrac12mR^2\dfrac vR \).
2. \( v_0=\dfrac32v\Rightarrow v=\dfrac23v_0 \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Answer: (c) \( \dfrac{v_0}{3\mu g} \)

Explanation:
1. Friction \( \mu mg \) decelerates the centre: \( v=v_0-\mu gt \). Pure rolling begins at \( v=\tfrac23v_0 \).
2. \( \dfrac13v_0=\mu gt_0\Rightarrow t_0=\dfrac{v_0}{3\mu g} \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Answer: (d) \( \cos^{-1}\dfrac47 \)

Explanation:
1. The cylinder pivots about the edge: \( I_{edge}=\tfrac32mR^2 \). Energy: \( mgR(1-\cos\theta)=\tfrac12\cdot\tfrac32mR^2\omega^2\Rightarrow R\omega^2\cdot R=\tfrac43gR(1-\cos\theta) \).
2. Contact is lost when \( N=0 \): \( mg\cos\theta=\dfrac{mv^2}R=\tfrac43mg(1-\cos\theta) \).
3. \( \cos\theta=\dfrac43-\dfrac43\cos\theta\Rightarrow\cos\theta=\dfrac47 \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Answer: (a) \( \sqrt{\dfrac{4gR}7} \)

Explanation:
1. \( v^2=\tfrac43gR(1-\cos\theta_c)=\tfrac43gR\cdot\dfrac37=\dfrac{4gR}7 \).
2. \( v=\sqrt{\dfrac{4gR}7} \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Answer: (b) 6

Explanation:
1. At leaving: translational KE \( =\tfrac12mv^2=\tfrac27mgR \), rotational KE \( =\tfrac14mv^2=\tfrac17mgR \) (constant in flight).
2. The centre falls \( R\cos\theta_c=\tfrac47R \) to the edge level, so translational KE \( =\tfrac27mgR+\tfrac47mgR=\tfrac67mgR \).
3. Ratio \( =\dfrac{6/7}{1/7}=6 \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Answer: (c) 1.63 N

Explanation:
1. Block (decelerating, magnitude \( a \)): \( mg\sin30^\circ-T=ma \). Drum: \( TR=\tfrac12MR^2\dfrac aR\Rightarrow T=\tfrac12Ma=a \).
2. \( 0.5(9.8)(0.5)=(0.5+1)a\Rightarrow a=1.633 m/s^{2} \), so \( T=1.63 N \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Answer: (d) 1.22 m

Explanation:
1. Speed of \( X =\omega R=10\times0.2=2 m/s \); deceleration \( a=1.633 m/s^{2} \) (from the tension calculation).
2. \( s=\dfrac{v^2}{2a}=\dfrac4{3.267}=1.22 m \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Answer: (a) \( \sqrt{\dfrac{14gR}{3}} \)

Explanation:
1. Mass in the rolled part is proportional to its cross-sectional area: at radius \( R/2 \) it is \( \dfrac M4 \), and the unrolled \( \dfrac{3M}4 \) lies on the floor at rest.
2. PE lost \( =MgR-\dfrac M4g\dfrac R2=\dfrac78MgR \). The rolled part has KE \( =\tfrac34\cdot\dfrac M4v^2=\dfrac{3Mv^2}{16} \) (rolling, \( I=\tfrac12mr^2 \)).
3. \( \dfrac{3v^2}{16}=\dfrac{7gR}8\Rightarrow v^2=\dfrac{14gR}3 \).

Teacher's Note:
1. For rolling without slipping use \( v = \omega R \) and take torques about the centre of mass or about the contact point.
2. Compare the kinetic energy split between translation and rotation using the factor \( 1 + I/(mR^2) \).

Answer: (b) \( -\dfrac{I_1\omega_1^2}{24} \)

Explanation:
1. \( \omega=\dfrac{I_1\omega_1+\tfrac{I_1}2\cdot\tfrac{\omega_1}2}{\tfrac32I_1}=\dfrac56\omega_1 \).
2. \( E_i=\tfrac12I_1\omega_1^2+\tfrac12\cdot\tfrac{I_1}2\cdot\tfrac{\omega_1^2}4=\tfrac9{16}I_1\omega_1^2 \); \( E_f=\tfrac12\cdot\tfrac32I_1\cdot\tfrac{25}{36}\omega_1^2=\tfrac{25}{48}I_1\omega_1^2 \).
3. \( E_f-E_i=\left(\tfrac{25}{48}-\tfrac{27}{48}\right)I_1\omega_1^2=-\tfrac{I_1\omega_1^2}{24} \).

Teacher's Note:
1. Angular momentum is conserved only about axes for which the net external torque is zero.
2. Write \( L = I\cdot \omega \) before and after the change and equate them, then compute the energy separately.

Answer: (c) \( \dfrac vL \)

Explanation:
1. About the centre of mass (rod midpoint): \( J\dfrac L2=I\omega \) with \( I=2M\left(\dfrac L2\right)^2=\dfrac{ML^2}2 \).
2. \( \omega=\dfrac{Mv\,L/2}{ML^2/2}=\dfrac vL \).

Teacher's Note:
1. Angular momentum is conserved only about axes for which the net external torque is zero.
2. Write \( L = I\cdot \omega \) before and after the change and equate them, then compute the energy separately.

Answer: (d) \( \omega_A=\omega \)

Explanation:
1. There is no friction, so no torque acts on either sphere during the collision and angular velocities do not change.
2. Hence \( A \) keeps its spin \( \omega \) (it stops translating) and \( B \), which was not spinning, stays at zero spin.

Teacher's Note:
1. Angular momentum is conserved only about axes for which the net external torque is zero.
2. Write \( L = I\cdot \omega \) before and after the change and equate them, then compute the energy separately.

Answer: (a) \( \dfrac{3v}{4a} \)

Explanation:
1. Angular momentum about \( O \) is conserved: before \( =Mv\dfrac a2 \) (the centre is at height \( a/2 \)).
2. After, \( I_O=\dfrac{Ma^2}6+M\left(\dfrac a{\sqrt2}\right)^2=\dfrac23Ma^2 \).
3. \( \omega=\dfrac{Mva/2}{\tfrac23Ma^2}=\dfrac{3v}{4a} \).

Teacher's Note:
1. Angular momentum is conserved only about axes for which the net external torque is zero.
2. Write \( L = I\cdot \omega \) before and after the change and equate them, then compute the energy separately.

Answer: (b) (a) and (c) only

Explanation:
1. Horizontal momentum before: \( 2(-1)+0.1(20)=0 \). The ball leaves with zero horizontal momentum, so the ring's centre of mass has zero velocity afterwards.
2. The ball's impulse still spins the ring (angular momentum about the ground point is conserved), so the ring rotates about its stationary centre: \( (a) \) true, \( (b) \) false.
3. The ring's contact point then slides over the ground, so kinetic friction acts and opposes this sliding, i.e. it acts to the left: \( (c) \) true, \( (d) \) false.

Teacher's Note:
1. Angular momentum is conserved only about axes for which the net external torque is zero.
2. Write \( L = I\cdot \omega \) before and after the change and equate them, then compute the energy separately.

Answer: (c) (a), (c) and (d) only

Explanation:
1. Momentum: \( 2m(v)-m(2v)=0 \) so \( v_c=0 \): \( (a) \) true.
2. Angular momentum about \( C \): \( 2mva+m(2v)(2a)=6mva \). \( I=\dfrac{8m(6a)^2}{12}+2ma^2+m(2a)^2=30ma^2 \).
3. \( \omega=\dfrac{6mva}{30ma^2}=\dfrac v{5a} \): \( (c) \) true, \( (b) \) false. \( E=\tfrac12(30ma^2)\dfrac{v^2}{25a^2}=\dfrac35mv^2 \): \( (d) \) true.

Teacher's Note:
1. Angular momentum is conserved only about axes for which the net external torque is zero.
2. Write \( L = I\cdot \omega \) before and after the change and equate them, then compute the energy separately.

Answer: (d) \( \dfrac{1}{4} \)

Explanation:
1. Linear: \( mv_0=MV \). Angular about the centre: \( mv_0\dfrac L2=\dfrac{ML^2}{12}\omega\Rightarrow\omega=\dfrac{6mv_0}{ML} \).
2. Energy: \( mv_0^2=MV^2+\dfrac{ML^2}{12}\omega^2=\dfrac{m^2v_0^2}M+\dfrac{3m^2v_0^2}M=\dfrac{4m^2v_0^2}M \).
3. \( \dfrac mM=\dfrac14 \).

Teacher's Note:
1. Angular momentum is conserved only about axes for which the net external torque is zero.
2. Write \( L = I\cdot \omega \) before and after the change and equate them, then compute the energy separately.

Answer: (a) \( \dfrac{2L}3 \)

Explanation:
1. \( V=\dfrac{mv_0}M=\dfrac{v_0}4 \) and \( \omega=\dfrac{6mv_0}{ML}=\dfrac{3v_0}{2L} \).
2. A point at distance \( x \) from the centre on the side opposite to \( A \) is at rest when \( V=\omega x\Rightarrow x=\dfrac{v_0/4}{3v_0/2L}=\dfrac L6 \).
3. \( AP=\dfrac L2+\dfrac L6=\dfrac{2L}3 \).

Teacher's Note:
1. Angular momentum is conserved only about axes for which the net external torque is zero.
2. Write \( L = I\cdot \omega \) before and after the change and equate them, then compute the energy separately.

Answer: (b) \( \dfrac{v_0}{2\sqrt2} \)

Explanation:
1. The angle turned is \( \omega t=\dfrac{3v_0}{2L}\cdot\dfrac{\pi L}{3v_0}=\dfrac\pi2 \).
2. \( P \) moves with the centre of mass at \( V=\dfrac{v_0}4 \), plus a rotational speed \( \omega\cdot\dfrac L6=\dfrac{v_0}4 \) about it. After \( \pi/2 \) these two velocities are perpendicular.
3. \( v_P=\dfrac{v_0}4\sqrt2=\dfrac{v_0}{2\sqrt2} \).

Teacher's Note:
1. Angular momentum is conserved only about axes for which the net external torque is zero.
2. Write \( L = I\cdot \omega \) before and after the change and equate them, then compute the energy separately.

Answer: (c) 6.3 m/s

Explanation:
1. \( I_P=\tfrac13(0.01)(0.6)^2+\left[\tfrac1{12}(0.02)(0.6)^2+0.02(0.9)^2\right]+0.05(1.2)^2=0.0012+0.0168+0.072=0.09 \) kg \( m^{2} \).
2. Angular momentum: \( 0.05v(1.2)=0.09\omega\Rightarrow\omega=\dfrac{2v}3 \). KE \( =\tfrac12(0.09)\dfrac{4v^2}9=0.02v^2 \).
3. PE gain \( =g(0.01\times0.3+0.02\times0.9+0.05\times1.2)=0.081g \). \( v^2=\dfrac{0.081(9.8)}{0.02}=39.7\Rightarrow v\approx6.3 m/s \).

Teacher's Note:
1. Angular momentum is conserved only about axes for which the net external torque is zero.
2. Write \( L = I\cdot \omega \) before and after the change and equate them, then compute the energy separately.

Answer: (d) 2.72 J

Explanation:
1. Momentum: \( 0.08(16)=(0.16+0.16)V\Rightarrow V=4 m/s \). Angular momentum about the centre: \( 0.08(10)(0.5)-0.08(6)(0.5)=0.16 \).
2. \( I=\dfrac{0.16(3)}{12}+2(0.08)(0.25)=0.08 \), so \( \omega=2 \) rad/s.
3. \( KE_i=\tfrac12(0.08)(100+36)=5.44 J \); \( KE_f=\tfrac12(0.32)(16)+\tfrac12(0.08)(4)=2.72 J \). Loss \( =2.72 J \).

Teacher's Note:
1. Angular momentum is conserved only about axes for which the net external torque is zero.
2. Write \( L = I\cdot \omega \) before and after the change and equate them, then compute the energy separately.

Answer: (a) T(x) is maximum at x = 0, falls parabolically with zero slope at x = 0 and becomes zero at x = l

Explanation:
1. The element beyond \( x \) has mass \( \lambda(l-x) \) with centre at \( \dfrac{l+x}2 \). \( T(x)=\lambda\omega^2\dfrac{(l-x)(l+x)}2=\dfrac{\lambda\omega^2}2(l^2-x^2) \).
2. This is an inverted parabola: maximum \( \dfrac{\lambda\omega^2l^2}2 \) at \( x = 0 \) with zero slope there, and zero at \( x = l \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (b) \( Gm\left[A\left(\dfrac1a-\dfrac1{a+L}\right)+BL\right] \)

Explanation:
1. \( dF=\dfrac{Gm\,(A+Bx^2)dx}{x^2} \).
2. \( F=Gm\left[A\left(\dfrac1a-\dfrac1{a+L}\right)+BL\right] \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.

Answer: (c) \( \dfrac\pi6 \)

Explanation:
1. \( \tau=rF\sin\theta\Rightarrow2.5=5(1)\sin\theta \).
2. \( \sin\theta=\tfrac12\Rightarrow\theta=\dfrac\pi6 \).

Teacher's Note:
1. Write the moment of inertia of each part about its own axis first, then shift it with the parallel-axis theorem.
2. Check the limiting cases (for example a sign or a factor of 2) to confirm the final expression.