JEE Physics System Of Particles and Rotational Motion MCQs Set 43

Answer: (d) 6

Explanation:
1. \( I=\dfrac{m\ell^2}{3}\sin^2\theta=\dfrac{m\ell^2}{3}\cdot\dfrac12=\dfrac{m\ell^2}{6} \).

Teacher's Note:
1. Only the component of distance perpendicular to the axis counts.
2. Each element at distance x along the rod is at x sin theta from the axis.

Answer: (b) 4

Explanation:
1. Positions: \( x_1=\ell/4 \) and \( x_2=\ell-\ell/6=5\ell/6 \); the centre of mass is at \( \ell/2 \).
2. Distances of the centre from the two heads: \( \ell/4 \) and \( \ell/3 \). The reaction is larger on the person nearer the centre of mass: \( N_1:N_2=\tfrac13:\tfrac14=4:3 \).

Teacher's Note:
1. The reaction at each support is inversely proportional to its distance from the centre of mass.
2. So n = 4.

Answer: (b) 4

Explanation:
1. Ring: friction torque \( \mu mgR \), \( I=mR^2 \), \( \alpha=\mu g/R \).
2. Disc (pressure uniform over the area): friction torque \( \int_0^R\mu\dfrac{mg}{\pi R^2}r\cdot2\pi r\,dr=\dfrac23\mu mgR \); \( I=\tfrac12mR^2 \); \( \alpha=\dfrac{4\mu g}{3R} \).
3. \( t=\omega/\alpha \): \( t_{ring}:t_{disc}=1:\tfrac34=4:3 \).

Teacher's Note:
1. The disc has its mass spread over smaller average distance than the ring.
2. Time to stop is inversely proportional to angular deceleration.

Answer: (d) 2

Explanation:
1. Torque \( =dL/dt=4\sqrt2 \) N m.
2. Perpendicular distance of the line \( x-y-4=0 \) from the origin: \( 4/\sqrt2=2\sqrt2 \) m.
3. \( F=\dfrac{4\sqrt2}{2\sqrt2}=2 \) N.

Teacher's Note:
1. Torque is force times its lever arm about the origin.
2. The line of motion is the line of action of the force.

Answer: (c) 8

Explanation:
1. If the centre of S moves at v, the top of S (and hence P) moves at 2v: \( KE_P=\tfrac12m(2v)^2=2mv^2 \).
2. \( KE_S=\tfrac12mv^2+\tfrac12\cdot\tfrac12mR^2\cdot\dfrac{v^2}{R^2}=\tfrac34mv^2 \).
3. Ratio \( =2:\tfrac34=8:3 \).

Teacher's Note:
1. The top of a rolling cylinder moves at twice the speed of its centre.
2. Remember the rotational energy of the cylinder.

Answer: (b) 5

Explanation:
1. Angular momentum about the contact point is conserved and its final value is zero: \( mv_0r-\tfrac25mr^2\omega_0=0 \).
2. \( \dfrac{\omega_0r}{v_0}=\dfrac52\Rightarrow\dfrac{2\omega_0r}{v_0}=5 \).

Teacher's Note:
1. Friction acts at the contact point, so angular momentum about it is conserved.
2. A solid sphere has I equal to two fifths m r squared.

Answer: (a) 2

Explanation:
1. Smooth spheres exchange translational velocities in an elastic head-on collision: B moves at \( v_0 \) and A stops translating; spins do not change.
2. \( KE_B=\tfrac12mv_0^2 \); \( KE_A=\tfrac12\cdot\tfrac23mR^2\cdot\dfrac{v_0^2}{R^2}=\tfrac13mv_0^2 \).
3. Ratio \( =\tfrac12:\tfrac13=3:2 \).

Teacher's Note:
1. Equal masses exchange velocities.
2. A hollow sphere has I equal to two thirds m R squared.

Answer: (d) 10

Explanation:
1. The cylinder stays on the floor only if \( F\le Mg \).
2. Torque about the contact point: \( F\cdot R=\tfrac32MR^2\alpha \), so \( \alpha=\dfrac{2F}{3MR} \).
3. \( \alpha_{max}=\dfrac{2g}{3R}=\dfrac{20}{2}=10 \) rad/s².

Teacher's Note:
1. The contact point is the instantaneous axis of rotation.
2. The upper limit of F is when the normal force becomes zero.

Answer: (d) 10

Explanation:
1. Total normal force \( =mg+F \); the total friction is limited to \( \mu(mg+F) \).
2. Torque about the contact line: \( FR=\tfrac32mR^2\alpha\Rightarrow a=R\alpha=\dfrac{2F}{3m} \). Friction needed \( f=ma=\dfrac{2F}{3} \).
3. \( \dfrac{2F}{3}\le\dfrac13(mg+F)\Rightarrow F\le mg=10 \) N.

Teacher's Note:
1. The pull increases the normal force and therefore the available friction.
2. Use the contact line as the axis.

Answer: (c) 20

Explanation:
1. From torque about the contact line: \( a=\dfrac{2F}{3m}=\dfrac{2\times10}{3}=\dfrac{20}{3} \) m/s².
2. So n = 20.

Teacher's Note:
1. The cylinder rolls about the contact line of the planks.
2. This uses the same relation as the maximum force.

Answer: (a) 12

Explanation:
1. Angular momentum about A is conserved. Translation contributes \( Mv_0R=20 \), spin contributes \( -\tfrac12MR^2\omega_0=-2 \) (opposite sense). Total \( L=18 \).
2. Moment of inertia about A: \( \tfrac12MR^2+MR^2=\tfrac32 \) kg m².
3. \( \omega=\dfrac{18}{1.5}=12 \) rad/s.

Teacher's Note:
1. The hinge force passes through A, so angular momentum about A is conserved.
2. Use the parallel axes theorem for the new axis.

Answer: (d) 9

Explanation:
1. Let the rod make angle \( \theta \) with the floor: the centre of mass is at \( \left(\tfrac\ell2\cos\theta,\tfrac\ell2\sin\theta\right) \) while the ends slide on the walls.
2. Kinetic energy \( =\tfrac12m\dfrac{\ell^2}{4}\dot\theta^2+\tfrac12\dfrac{m\ell^2}{12}\dot\theta^2=\dfrac{m\ell^2}{6}\dot\theta^2 \); potential energy \( =mg\dfrac\ell2\sin\theta \).
3. The equation of motion gives \( \dfrac{m\ell^2}{3}\ddot\theta=-mg\dfrac\ell2\cos\theta \): magnitude \( \alpha=\dfrac{3g\cos\theta}{2\ell}=\dfrac{3(10)(0.6)}{2}=9 \) rad/s².

Teacher's Note:
1. Both ends slide, so the hinge-type constraint is replaced by the geometric constraint.
2. Energy and Lagrange's method avoid the unknown wall and floor reactions.

Answer: (c) (C) and (D)

Explanation:
1. \( \vec v=\vec\omega\times\vec r \) and \( \vec a=\omega^2\vec r_\perp \) (for constant \( \omega \)) are equal only when the points coincide, so (A) and (B) are false.
2. Every particle at distance r from the axis has speed \( \omega r \): (C) true. All particles share the same angular velocity: (D) true.

Teacher's Note:
1. Velocity and acceleration are vectors, so the same speed does not mean the same velocity.
2. Angular speed is a property of the whole body.

Answer: (b) (A), (B) and (C)

Explanation:
1. Uniform rotation: \( \alpha=0 \) for all particles, particles on the axis do not move, and \( \omega \) is common: (A), (B), (C) true.
2. Linear speed \( \omega r\sin\phi \) depends on the particle's distance from the axis: (D) false.

Teacher's Note:
1. Uniform rotation means constant angular velocity.
2. The linear speed varies from zero at the poles to a maximum at the equator.

Answer: (a) (A), (B), (C) and (D)

Explanation:
1. (A) \( \alpha=\dfrac{mg(l/2)\sin60^\circ}{ml^2/3}=\dfrac{3\sqrt3g}{4l} \).
2. (B) Just after release \( \omega=0 \): along the rod \( N_r=mg\cos60^\circ=\dfrac{mg}2 \); perpendicular: \( N_t=mg\sin60^\circ-m\dfrac{\alpha l}{2}=\dfrac{\sqrt3mg}{8} \); \( N=mg\sqrt{\tfrac14+\tfrac3{64}}=\dfrac{\sqrt{19}mg}{8} \).
3. (C) \( mg\dfrac l2(1-\cos60^\circ)=\tfrac12\dfrac{ml^2}{3}\omega^2\Rightarrow\omega^2=\dfrac{3g}{2l} \).
4. (D) At the vertical \( N-mg=m\omega^2\dfrac l2=\dfrac34mg\Rightarrow N=\dfrac74mg \). All four are correct.

Teacher's Note:
1. Resolve the hinge force along and perpendicular to the rod.
2. The centre of mass has a centripetal acceleration once the rod is moving.

Answer: (c) (B) and (C)

Explanation:
1. Speeds: A (top) \( 2v \); B \( \sqrt2v \) (so B is slower than A and (A) is false); centre C \( v \).
2. In the frame of the centre every rim point has acceleration \( v^2/r \) towards the centre. At A the acceleration is perpendicular to the velocity: \( \rho=\dfrac{(2v)^2}{v^2/r}=4r \): (B) true.
3. At B the velocity makes 45° with the horizontal acceleration, so the normal component is \( \dfrac{v^2}{\sqrt2r} \): \( \rho=\dfrac{2v^2}{v^2/(\sqrt2r)}=2\sqrt2r \): (C) true. The centre moves in a straight line, so (D) is false.

Teacher's Note:
1. The acceleration of every point of a uniformly rolling wheel is v squared over r towards the centre.
2. Radius of curvature equals speed squared over the normal acceleration.

Answer: (c) (A), (B) and (C)

Explanation:
1. Pure rolling needs \( f=\tfrac27mg\sin\theta\le\mu mg\cos\theta \Rightarrow \mu\ge\tfrac27\tan\theta \): (A) true.
2. Static friction does no net work on a rolling body, so the whole loss of potential energy appears as kinetic energy: (B) true.
3. The net work of friction is zero, which is less than \( \mu mgS\cos\theta \) (and the friction needed is at most the limiting value): (C) true, (D) false.

Teacher's Note:
1. Friction needed for rolling is smaller than the limiting friction unless equality holds.
2. Rolling without slipping conserves mechanical energy.

Answer: (b) (A), (C) and (D)

Explanation:
1. The wall is smooth, so it changes only the velocity of the centre (reverses it) and not the spin: speed v is retained, (A) true.
2. The spin is then in the wrong sense for rolling in the new direction, so the cylinder slips, friction decelerates and reverses the spin (passing through zero), and pure rolling starts after some time: (C), (D) true, (B) false.

Teacher's Note:
1. A smooth wall exerts no torque about the centre.
2. Kinetic friction continues until the contact point stops slipping.

Answer: (d) (B) and (C)

Explanation:
1. The hinge exerts an impulsive force, so linear momentum is not conserved ((A) and (D) false).
2. That force has no moment about A, so angular momentum about A is conserved (B). The collision is elastic, so kinetic energy is conserved (C).

Teacher's Note:
1. The hinge force passes through the point about which torque is taken.
2. Elastic means no kinetic energy is lost.

Answer: (c) (A), (B) and (D)

Explanation:
1. Friction between the surfaces brings them to a common speed (A) and there is no external torque, so angular momentum is conserved (B).
2. \( \tfrac12MR^2\omega_0=(\tfrac12+1)MR^2\omega\Rightarrow\omega=\dfrac{\omega_0}{3} \): (D) true, (C) false.

Teacher's Note:
1. Moments of inertia: disc one half M R squared, ring M R squared.
2. Kinetic energy is lost to friction.

Answer: (a) (B) and (C)

Explanation:
1. Friction exerts a torque about the centre, so (A) is false. About the contact point angular momentum is conserved: initial \( mvR-\tfrac23mR^2\dfrac vR=\dfrac{mvR}{3} \).
2. When the spin is zero: \( mv'R=\dfrac{mvR}3\Rightarrow v'=\dfrac v3 \): (B) true.
3. Pure rolling: \( mv_fR+\tfrac23mRv_f=\dfrac{mvR}{3}\Rightarrow v_f=\dfrac v5 \): (C) true. Friction does work (kinetic energy is lost): (D) false.

Teacher's Note:
1. Use angular momentum about a point on the surface.
2. A hollow sphere has I equal to two thirds m R squared.

Answer: (d) (A) and (D)

Explanation:
1. Sliding begins when \( \tan\theta=\mu \); toppling begins when the vertical through the centre reaches the lower edge: \( \tan\theta=\dfrac ah \).
2. If \( \mu>a/h \) the block topples first (A); if \( \mu\lt a/h \) it slides first (D).

Teacher's Note:
1. The smaller of the two critical angles is reached first.
2. Compare tan theta for sliding and for toppling.

Answer: (c) (A), (C) and (D)

Explanation:
1. Net upward force \( T_1+T_2-Mg=2Mg-Mg=Mg \): acceleration g upward (A true, B false).
2. Torque about the centre \( =(T_2-T_1)R=MgR \): this equals the rate of change of angular momentum (D true).
3. \( \alpha=\dfrac{MgR}{\tfrac12MR^2}=\dfrac{2g}R \) (C true).

Teacher's Note:
1. The two tensions act at opposite sides of the centre.
2. Linear and angular motions are found separately.

Answer: (b) \( \dfrac{\omega_0}{3} \)

Explanation:
1. Angular momentum is conserved: \( \tfrac12mR^2\omega_0=\left[\tfrac12mR^2+2\cdot\dfrac m2R^2\right]\omega \).
2. \( \omega=\dfrac{\omega_0}{3} \).

Teacher's Note:
1. The balls start near the axis and end at distance R.
2. No external torque acts about the axis.