JEE Physics System Of Particles and Rotational Motion MCQs Set 50

Answer: (c) \( \omega=\frac{\omega_0}{1-\frac{a}{r_0}\theta} \)

Explanation:
1. The tension is perpendicular to the velocity and does no work, so the speed v of the particle is constant. Initially \( \omega_0=\frac{v}{r_0} \).
2. After turning through θ, the free length of cord is \( r=r_0-a\theta \), so \( \omega=\frac vr=\frac{v}{r_0-a\theta} \).
3. \( \omega=\frac{\omega_0}{1-a\theta/r_0} \)

Teacher's Note:
1. Constant speed with shrinking radius means increasing angular velocity.
2. Each radian of winding shortens the free cord by a.

Answer: (a) \( mk^2+m(a+b)^2-ma^2 \)

Explanation:
1. The axis x = a is at distance a from the centre of mass (taking the centre at the origin), so \( I_{cm}=mk^2-ma^2 \).
2. Shift to x = a + b, at distance a + b from the centre: \( I=I_{cm}+m(a+b)^2 \).
3. \( I=mk^2-ma^2+m(a+b)^2 \)

Teacher's Note:
1. The parallel-axis theorem only works through the centre of mass, so go to the centre first and then out.
2. Do not apply it directly between two non-central axes.

Answer: (b) 2 : 1

Explanation:
1. The top point of the cylinder moves at 2v, so the plank moves at 2v: \( KE_P=\tfrac12M(2v)^2=2Mv^2 \).
2. Hollow cylinder, rolling: \( KE_C=\tfrac12Mv^2+\tfrac12MR^2\left(\frac vR\right)^2=Mv^2 \).
3. \( KE_P:KE_C=2:1 \)

Teacher's Note:
1. The top of a rolling body moves at twice the centre speed.
2. A hollow cylinder has equal translational and rotational energy.

Answer: (c) \( 2\mu g \)

Explanation:
1. Kinetic friction is \( \mu mg \) on each body, in opposite directions (Newton's third law).
2. Plank: \( a_p=\mu g \) one way; sphere: \( a_s=\mu g \) the opposite way. The surface under the plank is smooth, so the plank is free to accelerate.
3. Relative acceleration \( =\mu g+\mu g=2\mu g \)

Teacher's Note:
1. Equal masses mean equal and opposite accelerations.
2. The plank is free to move, so do not treat it as fixed.

Answer: (d) zero

Explanation:
1. Take friction f forward. Translation: \( F+f=ma \).
2. Torque about the centre (ring, \( I=mR^2 \)) with rolling \( a=\alpha R \): \( FR-fR=mR^2\frac aR\Rightarrow F-f=ma \).
3. Subtracting the two equations: \( 2f=0\Rightarrow f=0 \)

Teacher's Note:
1. Pushing at the top of a ring makes the translation and rotation requirements agree automatically.
2. Always solve with both equations together.

Answer: (d) \( 2g\sin\theta \)

Explanation:
1. The cylinder is stationary, so the friction from the plank balances the component of its weight: \( f=mg\sin\theta \) up the slope.
2. Torque about its centre: \( fR=\tfrac12mR^2\alpha\Rightarrow\alpha=\frac{2g\sin\theta}{R} \).
3. No slipping means the plank's acceleration equals that of the contact point on the cylinder: \( a=\alpha R=2g\sin\theta \)

Teacher's Note:
1. Because the centre is fixed, the cylinder's contact point moves only by rotation.
2. Plank speed equals ωR at every instant.

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

Explanation:
1. Angular momentum about the ridge is conserved during the impact. Before: the centre of the cube is at height a/2 above O, so \( L=mv\frac a2 \).
2. After: the cube rotates about its edge, \( I_O=\frac{ma^2}{6}+m\left(\frac{a}{\sqrt2}\right)^2=\frac23ma^2 \).
3. \( \omega=\frac{mva/2}{2ma^2/3}=\frac{3v}{4a} \)

Teacher's Note:
1. Conserve angular momentum about the point where the impulsive force acts.
2. The distance from the cube's centre to a bottom edge is a/√2.

Answer: (a) \( v_0=\frac25v_1 \)

Explanation:
1. Take a fixed point O on the belt surface line. Friction acts along that line (zero torque about O), and gravity and the normal reaction are equal and opposite on the same vertical line (zero net torque), so angular momentum about O is conserved.
2. Initially \( L=mv_0r \). Finally the centre is at rest, and rolling on the moving belt requires the contact point to move at v1, i.e. \( \omega r=v_1 \).
3. Final \( L=I\omega=\tfrac25mr^2\cdot\frac{v_1}{r}=\tfrac25mrv_1 \), in the same sense as the initial one.
4. \( v_0=\frac25v_1 \)

Teacher's Note:
1. Friction acts along the line of the belt surface, so it has no torque about a point on that line.
2. At the end the sphere's centre is at rest, so only spin carries angular momentum.

Answer: (b) \( \frac{J}{2m} \)

Explanation:
1. Initially the ring slides with \( v_0=\frac Jm \) and no spin.
2. Friction acts at the contact point, so angular momentum about that point is conserved: \( mv_0R=mvR+mR^2\frac vR \).
3. \( mv_0R=2mvR\Rightarrow v=\frac{v_0}{2}=\frac{J}{2m} \)

Teacher's Note:
1. Use the contact point as the axis so that friction drops out.
2. For a ring the rolling condition splits the angular momentum equally between translation and spin.

Answer: (c) \( -4\ \text{kg m}^2/\text{s} \)

Explanation:
1. \( L_z=m(xv_y-yv_x) \).
2. With x = 1, y = 1, \( v_x=2 \), \( v_y=-2 \):
3. \( L_z=1\times(1\times(-2)-1\times2)=-4\ \text{kg m}^2/\text{s} \)

Teacher's Note:
1. Only the x and y components matter for the z-component of L.
2. The z-velocity gives no contribution to L_z.

Answer: (b) \( \cos\theta=\frac rR \)

Explanation:
1. For no rotation the torque of F about the point of contact must be zero, so the line of action of F must pass through the contact point.
2. The line is tangent to the drum, so its distance from the centre is r. The distance of the contact point from the line is \( R\cos\theta-r \).
3. \( R\cos\theta-r=0\Rightarrow\cos\theta=\frac rR \)

Teacher's Note:
1. Torque about the contact point removes friction and normal force from the equation.
2. For cos θ greater than r/R the spool rolls one way, otherwise the other way.

Answer: (d) x = 3

Explanation:
1. By the parallel-axis theorem \( I=I_{cm}+m(x-x_c)^2 \), which is smallest when the hinge passes through the centre of mass.
2. So the centre of mass is at the minimum of I: \( \frac{dI}{dx}=4x-12=0\Rightarrow x=3 \).
3. Here \( I_{cm}=I(3)=18-36+22=4 \) kg m² and the body's mass is 2 kg (the coefficient of x²)

Teacher's Note:
1. The moment of inertia about parallel axes is least for the axis through the centre of mass.
2. The coefficient of x² equals the body's mass.

Answer: (a) 75 J

Explanation:
1. Angular impulse about the hinge: \( J L=I\omega\Rightarrow10\times1=\tfrac13(2)(1)^2\omega \).
2. \( \omega=15\ \text{rad/s} \).
3. \( KE=\tfrac12I\omega^2=\tfrac12\cdot\tfrac23\cdot225=75\ \text{J} \)

Teacher's Note:
1. The hinge reaction is impulsive too, so take moments about the hinge.
2. KE can also be written as (JL)²/2I.

Answer: (b) \( \frac34 \)

Explanation:
1. Angular momentum about the hinge: \( mv\frac L2=\frac{ML^2}{3}\omega\Rightarrow\omega=\frac{3mv}{2ML} \).
2. Elastic collision, so kinetic energy is conserved: \( \tfrac12mv^2=\tfrac12\cdot\frac{ML^2}{3}\omega^2 \).
3. Substituting: \( mv^2=\frac{ML^2}{3}\cdot\frac{9m^2v^2}{4M^2L^2}=\frac{3m^2v^2}{4M}\Rightarrow\frac Mm=\frac34 \)

Teacher's Note:
1. Both conservation of angular momentum about the hinge and conservation of energy are needed.
2. Linear momentum is not conserved because the hinge exerts an impulse.

Answer: (c) \( 5v_0=2\omega_0R \)

Explanation:
1. Friction acts at the contact point, so angular momentum about the contact point is conserved.
2. Initially the translational part is \( mv_0R \) in one sense and the spin part is \( \tfrac25mR^2\omega_0 \) in the opposite sense.
3. Finally the sphere is at rest, so the total is zero: \( mv_0R=\tfrac25mR^2\omega_0\Rightarrow5v_0=2\omega_0R \)

Teacher's Note:
1. Use the contact point as the axis, so friction has no torque.
2. The two parts oppose each other because of the backspin.

Answer: (b) \( \frac l6 \)

Explanation:
1. Momentum: \( mv_0=mv_{cm}\Rightarrow v_{cm}=v_0 \) (the particle stops).
2. Angular momentum about O: \( mv_0x=\frac{ml^2}{12}\omega\Rightarrow\omega=\frac{12v_0x}{l^2} \).
3. End A is stationary when its velocity is zero: \( v_{cm}-\omega\frac l2=0\Rightarrow v_0=\frac{6v_0x}{l}\Rightarrow x=\frac l6 \)

Teacher's Note:
1. The far end moves backward by the rotation and forward by translation; set the two equal.
2. Take angular momentum about the centre of the rod.

Answer: (c) \( \frac53ml^2 \)

Explanation:
1. Rod 1 (contains O, which is its end): \( I_1=\frac13ml^2 \).
2. Rod 2 is perpendicular to rod 1 and attached to its far end. Its centre is at distance \( \sqrt{l^2+\frac{l^2}{4}} \) from O, so \( I_2=\frac{ml^2}{12}+m\cdot\frac{5l^2}{4}=\frac43ml^2 \).
3. \( I=\frac13ml^2+\frac43ml^2=\frac53ml^2 \)

Teacher's Note:
1. The second rod needs the parallel-axis theorem with the distance from its centre to O.
2. Use Pythagoras for that distance.

Answer: (a) \( v_{cm}=0,\ \omega=\frac{12v}{7l} \)

Explanation:
1. The momenta cancel, so \( v_{cm}=0 \).
2. Angular momentum about the centre: \( 2\cdot mv\frac l2=mvl \).
3. Moment of inertia of the system: \( \frac{ml^2}{12}+2m\left(\frac l2\right)^2=\frac{7}{12}ml^2 \).
4. \( \omega=\frac{mvl}{7ml^2/12}=\frac{12v}{7l} \)

Teacher's Note:
1. Particles on opposite ends add their angular momenta because they move in opposite directions.
2. Particles that stick become part of the rotating body.

Answer: (c) \( 5mv_0^2 \)

Explanation:
1. Ring: \( \tfrac12mv_0^2+\tfrac12mR^2\frac{v_0^2}{R^2}=mv_0^2 \).
2. Top particle moves at \( 2v_0 \): \( \tfrac12m(2v_0)^2=2mv_0^2 \). Bottom particle is at rest: 0.
3. Each side particle has speed \( \sqrt{v_0^2+v_0^2}=\sqrt2v_0 \): \( \tfrac12m(2v_0^2)=mv_0^2 \) each, so \( 2mv_0^2 \) for both.
4. Total: \( mv_0^2+2mv_0^2+0+2mv_0^2=5mv_0^2 \)

Teacher's Note:
1. Find each particle's speed from rolling: centre velocity plus the rotation velocity.
2. Side particles have the vector sum of v0 forward and v0 vertical.

Answer: (c) 3 and 4 only

Explanation:
1. Friction opposes the slipping, so it acts backward (4 true) and gives the sphere a spin.
2. Linear momentum: \( mv_0-ft=mv \). Angular momentum about the contact point: \( mv_0R=mvR+\tfrac25mR^2\frac vR \) at the end of slipping. Hence \( v=\frac57v_0 \), not zero (1 false).
3. Final spin: \( \omega=\frac vR=\frac{5v_0}{7R}<\frac{v_0}{R} \) (3 true, 2 false).

Teacher's Note:
1. Rolling starts once the contact point's velocity is zero, not when the sphere stops.
2. Final speed of a solid sphere in this case is 5/7 of the initial.

Answer: (b) 1 and 4 only

Explanation:
1. Normal acceleration \( \frac{v^2}{r}=\text{constant}\Rightarrow v\propto\sqrt r \).
2. As r increases, v increases (statement 1 true).
3. Angular speed \( \omega=\frac vr\propto\frac{\sqrt r}{r}=r^{-1/2} \) decreases (statement 4 true).

Teacher's Note:
1. Use v²/r = constant to link speed and radius.
2. Angular speed falls even though the linear speed rises.

Answer: (c) 3 and 4 only

Explanation:
1. Friction \( \mu mg \) gives \( a=\mu g \) and \( \alpha=\frac{2\mu g}{R} \). Rolling begins when \( \mu gt=R\omega_0-2\mu gt \).
2. Time \( t=\frac{R\omega_0}{3\mu g} \) and distance \( \tfrac12\mu gt^2 \) both contain μ.
3. Speed \( v=\mu gt=\frac{R\omega_0}{3} \) is independent of μ. Work done by friction equals the change in kinetic energy, \( \tfrac34mv^2-\tfrac14mR^2\omega_0^2 \), also independent of μ.

Teacher's Note:
1. Angular momentum about the contact point fixes the final speed without needing μ.
2. Work-energy theorem links the friction work to the speeds only.

Answer: (a) 1 and 2 only

Explanation:
1. Tangential acceleration \( a_t=\alpha l=kl\sin\theta \) (1 true, 3 false).
2. Use \( \omega\,d\omega=\alpha\,d\theta \): \( \frac{\omega^2}{2}=\int_0^\theta k\sin\theta\,d\theta=k(1-\cos\theta) \).
3. Centripetal acceleration \( \omega^2l=2kl(1-\cos\theta) \) (2 true, 4 false).

Teacher's Note:
1. Integrate ω dω = α dθ when α depends on angle.
2. Centripetal acceleration is ω² times the radius.

Answer: (b) \( \frac{mv^3}{4\sqrt2\,g} \)

Explanation:
1. At the highest point the velocity is horizontal: \( v_x=v\cos45^\circ=\frac v{\sqrt2} \).
2. Maximum height: \( H=\frac{v^2\sin^245^\circ}{2g}=\frac{v^2}{4g} \).
3. \( L=mv_xH=m\cdot\frac{v}{\sqrt2}\cdot\frac{v^2}{4g}=\frac{mv^3}{4\sqrt2\,g} \)

Teacher's Note:
1. At the top, L = m × horizontal speed × height (the perpendicular distance).
2. Do not use the range for this question.

Answer: (b) 2 and 3 only

Explanation:
1. Rolling: \( \vec v_A=0 \), \( \vec v_B=\vec v \), \( \vec v_C=2\vec v \) (all along the same direction).
2. (1) LHS = 2v, RHS = 2(v − 2v) = −2v: false. (2) LHS = v, RHS = v: true.
3. (3) LHS = 2v, RHS = 2v: true. (4) LHS = 2v, RHS = 4v: false.

Teacher's Note:
1. Write the three velocities in terms of v and substitute.
2. Vector statements must match in direction as well as magnitude.