JEE Physics System Of Particles and Rotational Motion MCQs Set 45

Answer: (d) \( \dfrac{\omega m}{m+2M} \)

Explanation:
1. Angular momentum is conserved: \( mR^2\omega=(m+2M)R^2\omega' \).
2. \( \omega'=\dfrac{m\omega}{m+2M} \).

Teacher's Note:
1. No external torque acts about the axis.
2. The two small masses add 2 M R squared to the moment of inertia.

Answer: (c) \( I_{AC}=I_{EF} \)

Explanation:
1. By symmetry, a square has the same moment of inertia about every axis through its centre in its plane: \( I_{AC}=I_{EF}=\dfrac{ml^2}{12} \).
2. By the parallel axes theorem \( I_{AD}=\dfrac{ml^2}{12}+m\dfrac{l^2}4=\dfrac{ml^2}3=4I_{EF} \), so \( I_{AD}\ne2I_{EF} \).

Teacher's Note:
1. For a square the principal moments in the plane are equal.
2. Use the parallel axes theorem for the side.

Answer: (b) \( \dfrac{g\sin\theta}{1+I/MR^2} \)

Explanation:
1. \( Mg\sin\theta-f=Ma \) and \( fR=I\dfrac aR \).
2. Hence \( a=\dfrac{g\sin\theta}{1+I/MR^2} \).

Teacher's Note:
1. The denominator is the ratio of the total effective inertia to the mass.
2. Check with a ring: I = M R squared gives g sin theta over 2.

Answer: (a) \( \dfrac23ma^2 \)

Explanation:
1. Through the centre: \( \dfrac{ma^2}{6} \).
2. Corner is at distance \( a/\sqrt2 \): \( I=\dfrac{ma^2}6+\dfrac{ma^2}2=\dfrac23ma^2 \).

Teacher's Note:
1. Use the parallel axes theorem.
2. The distance from the centre to a corner is half the diagonal.

Answer: (d) \( \dfrac16\dfrac{l^2\omega^2}{g} \)

Explanation:
1. The maximum speed is at the lowest position: \( \tfrac12\cdot\dfrac{ml^2}{3}\omega^2=mgh \).
2. \( h=\dfrac{l^2\omega^2}{6g} \).

Teacher's Note:
1. Energy at the bottom is purely rotational.
2. The moment of inertia about the end is m l squared over 3.

Answer: (b) \( \dfrac{r\omega_0}{2} \)

Explanation:
1. Angular momentum about the contact point is conserved: \( mr^2\omega_0=mvr+mr^2\omega \) with \( \omega=v/r \).
2. \( v=\dfrac{r\omega_0}{2} \).

Teacher's Note:
1. Friction acts at the point of contact.
2. Pure rolling begins when v equals omega r.

Answer: (a) more than 3 but less than 6

Explanation:
1. \( \tau=2F=40t-10t^2 \), \( \alpha=\dfrac\tau I=4t-t^2 \), \( \omega=2t^2-\dfrac{t^3}3 \) (from rest).
2. \( \omega=0 \) again at \( t=6 \) s. Angle \( \theta=\int_0^6\omega\,dt=\left[\dfrac{2t^3}3-\dfrac{t^4}{12}\right]_0^6=144-108=36 \) rad.
3. Rotations \( =\dfrac{36}{2\pi}=5.7 \).

Teacher's Note:
1. Integrate the angular acceleration twice.
2. The direction reverses when the angular velocity becomes zero again.

Answer: (d) first increases and then decreases

Explanation:
1. The system's angular momentum is conserved.
2. Moving towards the centre decreases I so the angular speed increases; moving away again increases I and the speed falls back to its original value.

Teacher's Note:
1. The product I omega stays constant.
2. The moment of inertia is least when the insect is at the centre.

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

Explanation:
1. \( I=M\left(\dfrac{R^2}4+\dfrac{l^2}{12}\right) \) with \( R^2=\dfrac{V}{\pi l} \): \( I=M\left(\dfrac V{4\pi l}+\dfrac{l^2}{12}\right) \).
2. \( \dfrac{dI}{dl}=0\Rightarrow-\dfrac V{4\pi l^2}+\dfrac l6=0\Rightarrow l^3=\dfrac{3V}{2\pi}=\dfrac32R^2l \).
3. \( \dfrac{l^2}{R^2}=\dfrac32\Rightarrow\dfrac lR=\sqrt{\dfrac32} \).

Teacher's Note:
1. Eliminate R using the fixed volume.
2. Differentiate with respect to the length and set the result to zero.

Answer: (a) \( \dfrac{181}{2}MR^2 \)

Explanation:
1. About O: central disc \( \tfrac12MR^2 \); each outer disc \( \tfrac12MR^2+M(2R)^2=\tfrac92MR^2 \). \( I_O=\tfrac12+6\times\tfrac92=\tfrac{55}{2}MR^2 \).
2. Total mass 7M, distance \( OP=3R \): \( I_P=I_O+7M(3R)^2=\tfrac{55}2MR^2+63MR^2 \).
3. \( I_P=\tfrac{181}{2}MR^2 \).

Teacher's Note:
1. Compute the moment of inertia about the central axis first.
2. Then shift by the parallel axes theorem with the whole mass.

Answer: (d) \( 4MR^2 \)

Explanation:
1. Full disc: \( \tfrac12(9M)R^2=\tfrac92MR^2 \).
2. Removed disc: \( \tfrac12M\left(\tfrac R3\right)^2+M\left(\tfrac{2R}3\right)^2=\tfrac1{18}MR^2+\tfrac49MR^2=\tfrac12MR^2 \).
3. \( I=\tfrac92MR^2-\tfrac12MR^2=4MR^2 \).

Teacher's Note:
1. Subtract the removed part from the full disc.
2. Use the parallel axes theorem for the removed disc.

Answer: (d) \( 8\pi \) rad/s²

Explanation:
1. \( \theta=50\times2\pi=100\pi \) rad in \( t=5 \) s, starting from rest: \( \theta=\tfrac12\alpha t^2 \).
2. \( \alpha=\dfrac{2\times100\pi}{25}=8\pi \) rad/s².

Teacher's Note:
1. Convert revolutions to radians first.
2. Starting from rest means the initial angular speed is zero.

Answer: (b) 20 s

Explanation:
1. Kinetic energy depends on \( \omega^2 \), so it is equal when \( \omega=-20 \) rad/s (rotation reversed).
2. \( \omega=\omega_0-\alpha t\Rightarrow -20=20-2t\Rightarrow t=20 \) s.

Teacher's Note:
1. After 10 s the body is momentarily at rest.
2. The torque keeps acting, so the body reverses its sense of rotation.

Answer: (a) \( \dfrac{ML^2}{12} \)

Explanation:
1. Perpendicular axes theorem for the two diagonals (equal by symmetry): \( I_z=I_{d1}+I_{d2}=2I_d \).
2. \( I_d=\dfrac12\cdot\dfrac{ML^2}{6}=\dfrac{ML^2}{12} \).

Teacher's Note:
1. The two diagonals are equivalent by symmetry.
2. The plate is planar, so the perpendicular axes theorem applies.

Answer: (d) \( MR^2\left(\dfrac12-\dfrac{16}{9\pi^2}\right) \)

Explanation:
1. About the centre O of the full circle, the half-disc has \( I_O=\tfrac12MR^2 \).
2. Its centre of mass is at \( d=\dfrac{4R}{3\pi} \) from O.
3. \( I_{cm}=I_O-Md^2=MR^2\left(\dfrac12-\dfrac{16}{9\pi^2}\right) \).

Teacher's Note:
1. Always shift to the centre of mass by subtracting M d squared.
2. The centre of mass of a semicircular plate is 4R over 3 pi from the centre.

Answer: (b) \( -14\hat i+10\hat j-9\hat k \)

Explanation:
1. Position vectors relative to (-1, 0, 1): \( \vec r_1=2\hat i+\hat j-\hat k \) and \( \vec r_2=\hat i+\hat j+\hat k \).
2. \( \vec r_1\times\vec F_1=(-11\hat i+10\hat j-12\hat k) \) and \( \vec r_2\times\vec F_2=(-3\hat i+0\hat j+3\hat k) \).
3. Sum: \( -14\hat i+10\hat j-9\hat k \).

Teacher's Note:
1. Measure position vectors from the given point, not the origin.
2. Torque is the sum of r cross F for each force.

Answer: (b) \( mg\ell\sin\theta \); zero at \( \theta=0^\circ \); maximum at \( \theta=90^\circ \)

Explanation:
1. Perpendicular distance of the line of the weight from the support is \( \ell\sin\theta \).
2. \( \tau=mg\ell\sin\theta \): zero at \( \theta=0^\circ \) (lowest point), largest (\( mg\ell \)) at \( \theta=90^\circ \).

Teacher's Note:
1. The lever arm is the horizontal distance of the bob from the vertical through the support.
2. The torque vanishes when the string is vertical.

Answer: (a) (a) \( mv^2\sin\alpha\cos\alpha \), (b) \( 2mv^2\sin\alpha\cos\alpha \)

Explanation:
1. Torque of the weight \( =mg\times \) horizontal distance from the point of projection.
2. At the highest point \( x=\dfrac{R}{2}=\dfrac{v^2\sin\alpha\cos\alpha}{g} \), so \( \tau=mv^2\sin\alpha\cos\alpha \).
3. On landing \( x=R=\dfrac{2v^2\sin\alpha\cos\alpha}{g} \), so \( \tau=2mv^2\sin\alpha\cos\alpha \).

Teacher's Note:
1. The weight is vertical, so only the horizontal distance matters.
2. Range is twice the horizontal distance to the highest point.

Answer: (d) 2.4 N and 2.6 N

Explanation:
1. Torques about the left end: \( T_R\times1=4\times0.5+1\times0.6=2.6 \) N, so \( T_R=2.6 \) N.
2. \( T_L+T_R=5 \) N gives \( T_L=2.4 \) N.

Teacher's Note:
1. Take torques about one end so the other tension drops out.
2. The object is nearer the right end, so the right string carries more.

Answer: (a) 3 rad/s²

Explanation:
1. Net torque \( =30 \) N m as before.
2. \( I=\dfrac{3\times2^2}{12}+9\times1^2=1+9=10 \) kg m².
3. \( \alpha=30/10=3 \) rad/s².

Teacher's Note:
1. Add the moment of inertia of the rod to that of the two masses.
2. The torque is unchanged because the rod's weight acts at the pivot.

Answer: (d) 39 N and 42 N

Explanation:
1. \( \alpha=3 \) rad/s², so each block has acceleration \( a=\alpha\times1=3 \) m/s².
2. The 3 kg block goes up: \( T=3(10+3)=39 \) N. The 6 kg block goes down: \( T=6(10-3)=42 \) N.

Teacher's Note:
1. The lighter block accelerates upward.
2. Tension exceeds weight for the block moving up.

Answer: (c) \( \dfrac{a\ell^4}{36} \)

Explanation:
1. Mass \( M=\int_0^\ell ax\,dx=\dfrac{a\ell^2}{2} \). Centre of mass: \( x_c=\dfrac{\int ax^2dx}{M}=\dfrac{2\ell}{3} \).
2. \( I_{left}=\dfrac{a\ell^4}{4} \), so \( I_{cm}=\dfrac{a\ell^4}{4}-M\left(\dfrac{2\ell}{3}\right)^2=a\ell^4\left(\dfrac14-\dfrac29\right)=\dfrac{a\ell^4}{36} \).

Teacher's Note:
1. The centre of mass is not at the middle for a non-uniform rod.
2. Use the parallel axes theorem from the left-end value.