JEE Physics System Of Particles and Rotational Motion MCQs Set 41

Answer: (b) \( \dfrac{\pi mR}{2I} \)

Explanation:
1. Linear: \( mv=I\Rightarrow v=I/m \). Angular impulse about the centre: \( Ih=\tfrac25mR^2\omega\Rightarrow\omega=\dfrac{I(4R/5)}{\tfrac25mR^2}=\dfrac{2I}{mR} \).
2. The surface is smooth, so both stay constant. B touches the ground after half a turn: \( t=\pi/\omega=\dfrac{\pi mR}{2I} \).

Teacher's Note:
1. On a smooth floor no friction acts, so the spin and the velocity stay constant.
2. The top point reaches the bottom after rotating by pi.

Answer: (a) \( \dfrac{\pi R}{2} \)

Explanation:
1. \( v=I/m \), \( \omega=2I/mR \), time for half a turn \( t=\pi/\omega=\dfrac{\pi mR}{2I} \).
2. Displacement \( =vt=\dfrac Im\cdot\dfrac{\pi mR}{2I}=\dfrac{\pi R}{2} \).

Teacher's Note:
1. Speed and spin are both independent of time here.
2. The answer does not depend on the impulse.

Answer: (c) \( \dfrac{a\tan\theta}{2} \)

Explanation:
1. Constant velocity: \( N=mg\cos\theta \), \( f=mg\sin\theta \) (acting on the lower face, a/2 below the centre).
2. No rotation: torques about the centre cancel: \( Nx=f\cdot\dfrac a2 \).
3. \( x=\dfrac{a}{2}\tan\theta \).

Teacher's Note:
1. The normal force is shifted ahead of the centre to balance the friction torque.
2. The same relation gives the toppling angle when x reaches a/2.

Answer: (c) 250

Explanation:
1. Initial speed \( =3000/60=50 \) rev/s; average \( =25 \) rev/s.
2. Revolutions \( =25\times10=250 \).

Teacher's Note:
1. Convert rpm to revolutions per second.
2. Uniform deceleration means the average speed is the mean of start and end.

Answer: (a) \( I_A\lt I_B \)

Explanation:
1. \( I=\tfrac12MR^2=\tfrac12(\rho\pi R^2t)R^2\propto R^4t \).
2. \( I_B/I_A=(2)^4\cdot\tfrac12=8 \), so \( I_A\lt I_B \).

Teacher's Note:
1. Mass is proportional to R squared times thickness.
2. So I goes as R to the fourth power times t.

Answer: (c) If the axes are parallel, \( I_1\lt I_2 \)

Explanation:
1. Parallel axes theorem: \( I_2=I_1+Md^2>I_1 \) when axis 2 is parallel to axis 1 and displaced.
2. For non-parallel axes no such relation holds in general, and \( I_1\lt I_2 \) does not imply parallel axes.

Teacher's Note:
1. The theorem needs the two axes to be parallel.
2. The centre-of-mass axis gives the minimum moment of inertia among parallel axes.

Answer: (b) \( \dfrac{m(r^2+d^2)}{4} \), which is less than \( \dfrac{mr^2}{2} \)

Explanation:
1. By the perpendicular axes theorem \( I_z=I_x+I_y=\dfrac{mr^2}{4}+\dfrac{md^2}{4} \).
2. Since \( d\lt r \), \( I_z\lt \dfrac{mr^2}{2} \).

Teacher's Note:
1. Use the perpendicular axes theorem with the two principal axes.
2. A circle of radius r would give mr squared over 2.

Answer: (d) \( \dfrac{5Ma^2}{18} \)

Explanation:
1. Each face has mass M/6. The two faces perpendicular to the axis: \( 2\times\dfrac{M}{6}\cdot\dfrac{a^2}{6}=\dfrac{Ma^2}{18} \).
2. Each of the four other faces: \( \dfrac M6\left(\dfrac{a^2}{12}+\dfrac{a^2}{4}\right)=\dfrac{Ma^2}{18} \); four of them give \( \dfrac{4Ma^2}{18} \).
3. Total \( \dfrac{5Ma^2}{18} \).

Teacher's Note:
1. A square plate has I = m a squared over 6 about a perpendicular axis through its centre.
2. Use the parallel axes theorem for the side faces.

Answer: (c) \( \dfrac{ma^2(10+3\pi)}{3} \)

Explanation:
1. Linear density \( =m/a \). Each side: mass m, \( I=\dfrac{ma^2}{12}+m\dfrac{a^2}{4}=\dfrac{ma^2}{3} \); four sides: \( \dfrac43ma^2 \).
2. Each semicircle: mass \( \pi m/2 \), radius a/2, centre of mass \( 2r/\pi=a/\pi \) from its diameter, so at distance \( a/2+a/\pi \) from the centre. \( I=M\left[r^2-\dfrac{a^2}{\pi^2}+\left(\dfrac a2+\dfrac a\pi\right)^2\right]=ma^2\left(\dfrac\pi4+\dfrac12\right) \). Four: \( (\pi+2)ma^2 \).
3. Total \( \left(\dfrac{10}{3}+\pi\right)ma^2 \).

Teacher's Note:
1. Treat the square sides and the semicircles separately and add.
2. Remember the parallel axes theorem uses the centre of mass of each semicircle.

Answer: (b) \( 62\hat k \)

Explanation:
1. \( \vec\tau=\vec r\times\vec F=(xF_y-yF_x)\hat k=[(-5)(-10)-(-3)(4)]\hat k=62\hat k \).

Teacher's Note:
1. Use the determinant form of the cross product.
2. Be careful with the two minus signs.

Answer: (a) remain constant

Explanation:
1. The torque of a couple equals (force) times (perpendicular distance between the forces) and does not depend on the chosen axis.

Teacher's Note:
1. A couple has the same moment about every point.
2. The net force of a couple is zero.

Answer: (c) \( \tfrac13 \)

Explanation:
1. Horizontal: \( N_2=\mu N_1 \). Vertical: \( N_1+\mu N_2=mg\Rightarrow N_1=\dfrac{mg}{1+\mu^2} \).
2. Torque about the foot: \( N_2(4)+\mu N_2(3)=mg(1.5) \Rightarrow \mu N_1(4+3\mu)=1.5mg \).
3. \( \dfrac{\mu(4+3\mu)}{1+\mu^2}=1.5\Rightarrow3\mu^2+8\mu-3=0\Rightarrow\mu=\tfrac13 \).

Teacher's Note:
1. Take torques about the foot of the ladder to avoid the floor forces.
2. Friction at the wall acts vertically.

Answer: (c) \( N_A=w\left(1-\dfrac xd\right),\ N_B=\dfrac{wx}{d} \)

Explanation:
1. Torque about A: \( N_Bd=wx\Rightarrow N_B=wx/d \).
2. \( N_A=w-N_B=w(1-x/d) \).

Teacher's Note:
1. Take torques about one support.
2. The two reactions add up to the weight.

Answer: (b) \( \sqrt{W_1W_2} \)

Explanation:
1. Arms of lengths \( l_1,l_2 \): \( Wl_1=W_1l_2 \) and \( W_2l_1=Wl_2 \).
2. Multiplying: \( W^2=W_1W_2\Rightarrow W=\sqrt{W_1W_2} \).

Teacher's Note:
1. The balance arms are unequal.
2. Multiply the two balance conditions.

Answer: (a) \( N_1=N_2 \)

Explanation:
1. Both normals are perpendicular to the rod, and the weight acts at the centre.
2. Torque about the centre: \( N_1\cdot\dfrac l4=N_2\cdot\dfrac l4\Rightarrow N_1=N_2 \).

Teacher's Note:
1. The contact points are equally far from the centre along the rod.
2. Normal force at a point contact on an edge is perpendicular to the rod.

Answer: (d) \( \omega_A>\omega_B \)

Explanation:
1. \( I\propto R^4t \): \( I_B/I_A=4^4\cdot\tfrac14=64 \).
2. \( \alpha=\tau/I \) so \( \alpha_A=64\alpha_B \), hence \( \omega_A>\omega_B \).

Teacher's Note:
1. Angular acceleration is torque divided by moment of inertia.
2. The ratio of speeds does not depend on the torque.

Answer: (c) \( \tfrac12m\omega^2l \)

Explanation:
1. The centre of mass moves in a circle of radius l/2: the hinge supplies the centripetal force \( m\omega^2\dfrac l2 \).

Teacher's Note:
1. Treat the rod as a point mass at its centre of mass for the net force.
2. The weight is balanced by the vertical normal force from the surface.

Answer: (b) \( \dfrac{15g}{16} \)

Explanation:
1. \( \alpha=\dfrac{mg(L/2)}{mL^2/3}=\dfrac{3g}{2L}=\dfrac{3g}{3.2}=\dfrac{15g}{16} \).

Teacher's Note:
1. The weight acts at the centre, a distance L/2 from the pivot.
2. The moment of inertia about the end is mL squared over 3.

Answer: (a) decrease

Explanation:
1. Before: \( mg/2 \). After: the rod rotates about the other end, \( \alpha=\dfrac{3g}{2L} \), \( a_{cm}=\alpha\dfrac L2=\dfrac{3g}{4} \).
2. \( mg-N=ma_{cm}\Rightarrow N=\dfrac{mg}{4} \), which is less than \( mg/2 \).

Teacher's Note:
1. Use the rotation about the supported end just after release.
2. Newton's third law makes the two forces always equal.

Answer: (c) \( \sqrt{\dfrac{24g}{7\ell}} \)

Explanation:
1. \( I=\dfrac{m\ell^2}{12}+m\left(\dfrac\ell4\right)^2=\dfrac{7m\ell^2}{48} \).
2. The centre falls by \( \ell/4 \): \( mg\dfrac\ell4=\tfrac12\cdot\dfrac{7m\ell^2}{48}\omega^2 \).
3. \( \omega^2=\dfrac{24g}{7\ell} \).

Teacher's Note:
1. Use the parallel axes theorem for the hinge position.
2. The drop of the centre of mass equals its distance from the hinge.

Answer: (b) \( 3A_0/4 \)

Explanation:
1. Torque is the rate of change of angular momentum: \( \tau=\dfrac{4A_0-A_0}{4}=\dfrac{3A_0}{4} \).

Teacher's Note:
1. Use torque equals change in angular momentum over time.
2. The size of the wheel is not needed.

Answer: (b) remains constant

Explanation:
1. The magnitude is \( mvx_0 \), where \( x_0 \) is the fixed perpendicular distance from the origin to the line of motion.
2. No torque acts, so \( \vec L \) is constant.

Teacher's Note:
1. Angular momentum is constant when there is no external torque.
2. The perpendicular distance stays the same.

Answer: (c) remains unchanged

Explanation:
1. There is no external torque about the axis, so angular momentum is conserved.
2. The moment of inertia increases and the angular speed decreases, but \( L=I\omega \) stays the same.

Teacher's Note:
1. Internal changes of the body do not alter its angular momentum.
2. The angular velocity falls to compensate the larger moment of inertia.

Answer: (b) \( 2\ell \)

Explanation:
1. The string leaves the top point, which moves at speed 2v when the centre moves at v. The string passes the boy's hand at this speed 2v.
2. The cylinder takes \( t=\ell/v \) to reach him, so the length is \( 2vt=2\ell \).

Teacher's Note:
1. The top of a rolling cylinder moves at twice the speed of its centre.
2. The length of string drawn equals the distance travelled by the top of the cylinder.

Answer: (a) \( \dfrac{2Mg\sin\theta}{K} \)

Explanation:
1. Friction does no work in pure rolling, so mechanical energy is conserved.
2. At the lowest point the speed is zero: \( Mgx\sin\theta=\tfrac12Kx^2 \).
3. \( x=\dfrac{2Mg\sin\theta}{K} \).

Teacher's Note:
1. At maximum displacement both translational and rotational kinetic energies are zero.
2. Compare with the equilibrium extension, which is half this value.