JEE Physics System Of Particles and Rotational Motion MCQs Set 46

Answer: (b) \( \dfrac{a\cos\alpha\sin^2\alpha}{2b-a\cos^2\alpha\sin\alpha} \)

Explanation:
1. The wall reaction N is perpendicular to the rod and acts at distance \( b/\sin\alpha \) from the foot. Torque about the foot: \( N\dfrac{b}{\sin\alpha}=mg\dfrac a2\cos\alpha\Rightarrow N=\dfrac{mga\sin\alpha\cos\alpha}{2b} \).
2. Horizontal: \( f=N\sin\alpha \). Vertical: floor reaction \( R=mg-N\cos\alpha \).
3. \( \mu=\dfrac fR=\dfrac{a\cos\alpha\sin^2\alpha}{2b-a\cos^2\alpha\sin\alpha} \).

Teacher's Note:
1. The wall exerts a force perpendicular to the rod at the contact point.
2. At the point of slipping friction is mu times the floor reaction.

Answer: (c) \( y=\dfrac{\sqrt{2ax}}{\omega} \) (a parabola)

Explanation:
1. \( v=at \), \( x=\tfrac12at^2 \), so \( t=\sqrt{2x/a} \) and \( y=\dfrac v\omega=\dfrac{at}{\omega} \).
2. \( y=\dfrac{a}{\omega}\sqrt{\dfrac{2x}{a}}=\dfrac{\sqrt{2ax}}{\omega} \).

Teacher's Note:
1. The distance of the instantaneous axis from the centre is v over omega.
2. Eliminate t using x equal to half a t squared.

Answer: (d) \( w_1=\dfrac{F}{m_1+\tfrac27m_2},\ w_2=\tfrac27w_1 \)

Explanation:
1. Friction f: \( F-f=m_1w_1 \), \( f=m_2w_2 \), \( fR=\tfrac25m_2R^2\alpha\Rightarrow\alpha R=\tfrac52w_2 \).
2. No slipping: \( w_1=w_2+\alpha R=\tfrac72w_2\Rightarrow w_2=\tfrac27w_1 \).
3. \( F=m_1w_1+m_2w_2=\left(m_1+\tfrac27m_2\right)w_1 \).

Teacher's Note:
1. The contact point of the sphere moves with the plank.
2. Write the equation of each body, then the rolling condition.

Answer: (a) \( \dfrac{125}{12}\approx10.4 \) rad/s

Explanation:
1. The bottom of the pipe moves with the plate: \( v_p=v_c+\omega r \).
2. \( \omega=\dfrac{5/3-5/6}{0.08}=\dfrac{125}{12}\approx10.4 \) rad/s (r is the radius, 0.08 m).

Teacher's Note:
1. Use the radius, not the diameter.
2. The plate is faster than the centre, so the pipe spins counter-clockwise.

Answer: (d) \( \dfrac{mg}{6} \), up the incline

Explanation:
1. Rolling without slipping with friction f up the plane: \( mg\sin30^\circ-f=ma \) and \( fR=\tfrac12mR^2\alpha,\ a=\alpha R\Rightarrow f=\tfrac12ma \).
2. \( \tfrac12mg=\tfrac32ma\Rightarrow a=\dfrac g3 \) (retardation), \( f=\dfrac{mg}6 \) directed up the incline.

Teacher's Note:
1. The disc is slowing down in translation and spinning down in rotation, which needs a torque against the spin.
2. Friction does not have to be at its limiting value in rolling.

Answer: (d) \( \dfrac{2v_0}{3} \) and \( \dfrac{v_0}{3\mu g} \)

Explanation:
1. \( v=v_0-\mu gt \) and \( \omega=\dfrac{\mu mgR}{\tfrac12mR^2}t=\dfrac{2\mu g}{R}t \).
2. Rolling when \( v=\omega R \): \( v_0-\mu gt=2\mu gt\Rightarrow t_0=\dfrac{v_0}{3\mu g} \).
3. \( v=v_0-\dfrac{v_0}3=\dfrac{2v_0}3 \).

Teacher's Note:
1. Rolling starts when v equals omega R.
2. Friction decelerates the centre and accelerates the spin.

Answer: (c) \( -\dfrac16mv_0^2 \)

Explanation:
1. Final rolling speed \( \tfrac23v_0 \). Final kinetic energy \( =\tfrac12\cdot\tfrac32m\left(\tfrac{2v_0}3\right)^2=\tfrac13mv_0^2 \).
2. Work by friction \( =\Delta K=\tfrac13mv_0^2-\tfrac12mv_0^2=-\tfrac16mv_0^2 \).

Teacher's Note:
1. Use the work-energy theorem.
2. The rolling kinetic energy of a disc is three quarters m v squared.

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

Explanation:
1. Rolled mass at radius r is \( M\dfrac{r^2}{R^2} \); at \( r=R/2 \) it is \( M/4 \). The flat unrolled part has no height and does not move.
2. Energy: \( MgR=\dfrac M4g\dfrac R2+\tfrac34\cdot\dfrac M4v^2 \) (rolling cylinder kinetic energy \( \tfrac34m'v^2 \)).
3. \( \dfrac78gR=\dfrac3{16}v^2\Rightarrow v=\sqrt{\dfrac{14gR}{3}} \).

Teacher's Note:
1. Cross-section area is conserved: pi r squared is proportional to the rolled length.
2. The unrolled carpet lies still on the floor.

Answer: (a) \( \dfrac{3W}{8} \)

Explanation:
1. At the limit, friction at floor \( f_1=0.5N_1 \), at wall \( f_2=0.5N_2 \), and horizontally \( N_2=f_1 \), so \( N_2=0.5N_1,\ f_2=0.25N_1 \).
2. Torque about the centre: \( F=f_1+f_2=0.75N_1 \).
3. Vertical: \( N_1+F+f_2=W\Rightarrow2N_1=W \), so \( N_1=\dfrac W2 \) and \( F=\dfrac{3W}8 \).

Teacher's Note:
1. Take torques about the axis of the cylinder.
2. At the verge of rotating both friction forces are at their limits.

Answer: (a) 6.3 m/s

Explanation:
1. \( I_P=\tfrac13(0.01)(0.6)^2+0.02\left[\tfrac{0.6^2}{12}+0.9^2\right]+0.05(1.2)^2=0.09 \) kg m².
2. To rise to the horizontal each mass gains height equal to its distance from P: \( g(0.01\times0.3+0.02\times0.9+0.05\times1.2)=0.081g \) J. \( \tfrac12I\omega^2=0.081g\Rightarrow\omega=4.2 \) rad/s.
3. Angular momentum about P: \( 0.05\,v\,(1.2)=0.09\times4.2\Rightarrow v=6.3 \) m/s.

Teacher's Note:
1. Angular momentum about the pivot is conserved in the collision.
2. Energy is conserved after the collision.

Answer: (d) \( \dfrac23mg \)

Explanation:
1. Tipping about the lower far edge: \( F\cdot\dfrac{3a}4=mg\cdot\dfrac a2 \).
2. \( F=\dfrac23mg \).

Teacher's Note:
1. Take torques about the edge that the cube pivots on.
2. The weight acts a/2 from that edge.

Answer: (c) \( \sqrt3\,m\omega^2\ell \)

Explanation:
1. The hinge supplies the centripetal force for the centre of mass of the body (mass 3m).
2. The centroid is at distance \( \ell/\sqrt3 \) from A: \( F=3m\omega^2\dfrac{\ell}{\sqrt3}=\sqrt3m\omega^2\ell \).

Teacher's Note:
1. Use the centre of mass: the net force is total mass times centripetal acceleration of the centre of mass.
2. Distance of the centroid from a vertex is the side over root 3.

Answer: (b) 6 N

Explanation:
1. Treat the object as one body: \( M=4 \) kg, \( I=2\times\tfrac12(2)(0.1)^2=0.02 \) kg m².
2. Total friction f: \( a_{cm}=f/M \), \( \alpha=fR/I \). Rolling relative to the truck: \( a_{truck}=a_{cm}+\alpha R \), so \( 9=f\left(\dfrac14+\dfrac{0.01}{0.02}\right)=0.75f \).
3. \( f=12 \) N in total, i.e. 6 N on each disc.

Teacher's Note:
1. The friction on the whole object is shared equally by the two discs.
2. Compare the acceleration of the contact point with the truck acceleration.

Answer: (b) \( \cos^{-1}\dfrac47 \) and \( \sqrt{\dfrac{4}{7}gr} \)

Explanation:
1. Rolling about the edge: \( I=\tfrac32mr^2 \), so kinetic energy \( =\tfrac34mv^2=mgr(1-\cos\theta) \).
2. Contact is lost when the normal reaction vanishes: \( mg\cos\theta=\dfrac{mv^2}r\Rightarrow v^2=gr\cos\theta \).
3. \( \tfrac34\cos\theta=1-\cos\theta\Rightarrow\cos\theta=\dfrac47 \) and \( v=\sqrt{\dfrac{4gr}7} \).

Teacher's Note:
1. The centre of mass moves on a circle of radius r about the edge.
2. The cylinder leaves when the normal force is zero.

Answer: (a) 6

Explanation:
1. After leaving, the angular velocity is constant, so \( K_R=\tfrac12\cdot\tfrac12mr^2\left(\dfrac vr\right)^2=\tfrac14mv^2=\dfrac{mgr}7 \).
2. The centre of mass has dropped by r from the start: total \( K=mgr \).
3. \( K_T=mgr-\dfrac{mgr}{7}=\dfrac{6mgr}7 \), so \( K_T/K_R=6 \).

Teacher's Note:
1. Rotation does not change once contact is lost.
2. Use energy conservation for the total kinetic energy.

Answer: (d) \( 2\pi\left(\dfrac{\alpha R^4}{4}+\dfrac{\beta R^5}{5}\right) \)

Explanation:
1. Ring element: \( dm=(\alpha+\beta x)2\pi x\,dx \).
2. \( I=\int_0^Rx^2dm=2\pi\int_0^R(\alpha x^3+\beta x^4)dx=2\pi\left(\dfrac{\alpha R^4}4+\dfrac{\beta R^5}5\right) \).

Teacher's Note:
1. Divide the disc into thin rings.
2. The mass of a ring is the density at that radius times its area.

Answer: (c) \( \dfrac{3}{10}mR^2 \)

Explanation:
1. \( I=\int_0^h\tfrac12\left(\dfrac{Rz}h\right)^2\cdot\dfrac{3m}{h^3}z^2dz \).
2. \( I=\dfrac{3mR^2}{2h^5}\cdot\dfrac{h^5}5=\dfrac3{10}mR^2 \).

Teacher's Note:
1. The mass of a disc slice is proportional to z squared.
2. Add the moments of inertia of the discs about the axis.

Answer: (c) \( \vec N=(aB-bA)\hat k \); arm \( \ell=\dfrac{|aB-bA|}{\sqrt{A^2+B^2}} \)

Explanation:
1. \( \vec N=\vec r\times\vec F=(aB-bA)\hat k \).
2. Arm \( \ell=\dfrac{|N|}{|F|}=\dfrac{|aB-bA|}{\sqrt{A^2+B^2}} \).

Teacher's Note:
1. The cross product of two vectors in the x-y plane points along z.
2. Torque magnitude equals force times arm.

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

Explanation:
1. No impulsive hinge force requires the impulse to act at the centre of percussion.
2. \( x=\dfrac{I}{M\,\ell/2}=\dfrac{M\ell^2/3}{M\ell/2}=\dfrac{2\ell}3 \).

Teacher's Note:
1. The hinge reaction is zero when the impulse acts at the centre of percussion.
2. The centre of percussion of a rod hinged at one end is 2 l over 3 from the hinge.

Answer: (c) 4.0 m

Explanation:
1. The path between two contacts is one arch of a cycloid.
2. Its length is \( 8R=8\times0.5=4.0 \) m.

Teacher's Note:
1. One arch of a cycloid has length 8R.
2. This is larger than the circumference.

Answer: (b) \( \dfrac{7}{10}mv^2\left(1+\dfrac{2r^2}{7R^2}\right) \)

Explanation:
1. Spin about the axle \( \omega_1=v/r \); precession about the vertical axis \( \omega_2=v/R \) (perpendicular components).
2. \( K=\tfrac12mv^2+\tfrac12\cdot\tfrac25mr^2(\omega_1^2+\omega_2^2)=\tfrac12mv^2+\tfrac15mv^2\left(1+\dfrac{r^2}{R^2}\right) \).
3. \( K=\dfrac7{10}mv^2+\dfrac15mv^2\dfrac{r^2}{R^2}=\dfrac7{10}mv^2\left(1+\dfrac{2r^2}{7R^2}\right) \).

Teacher's Note:
1. The two angular velocity components are perpendicular, so their squares add.
2. Add translational kinetic energy of the centre.

Answer: (a) \( \dfrac{v_1}{L}\cdot\dfrac{12\sin\beta}{1+3\sin^2\beta} \)

Explanation:
1. Vertical impulse J at A; horizontal offset of A from the centre of mass \( d=\tfrac L2\sin\beta \). \( J=m(v_c'+v_1) \) and \( Jd=\tfrac{mL^2}{12}\omega \).
2. Elastic: the vertical velocity of A reverses, \( v_c'+\omega d=v_1 \).
3. Solving: \( \omega=\dfrac{12v_1\sin\beta}{L(1+3\sin^2\beta)} \) (verified symbolically).

Teacher's Note:
1. The impulse at A produces both linear and angular momentum.
2. Use e = 1 for the velocity of the contact point.