JEE Physics System Of Particles and Rotational Motion MCQs Set 35

Answer: (b) \( \frac{17}{7}mg \)

Explanation:
1. Energy: \( mgR=\frac12mv^2\left(1+\frac25\right)\Rightarrow v^2=\frac{10gR}{7} \).
2. \( N-mg=\frac{mv^2}{R}=\frac{10mg}7\Rightarrow N=\frac{17}{7}mg \)

Teacher's Note:
1. Use energy conservation with translational plus rotational energy, then add the centripetal requirement.

Answer: (b) The cylinder has a greater translational speed than the hoop

Explanation:
1. \( K=\frac12mv^2(1+k) \): hoop \( k=1 \), cylinder \( k=\frac12 \).
2. For equal K, the body with the smaller factor moves faster: the cylinder.

Teacher's Note:
1. Rolling kinetic energy is (1 + k) times the translational part.

Answer: (c) 3 : 2

Explanation:
1. Equal masses, elastic head-on collision: the translational velocities are exchanged, and smooth surfaces mean A keeps its spin.
2. B: \( K_B=\frac12mV_0^2 \). A: only rotation remains, \( K_A=\frac12\cdot\frac23mR^2\left(\frac{V_0}{R}\right)^2=\frac13mV_0^2 \).
3. \( \frac{K_B}{K_A}=\frac{1/2}{1/3}=\frac32 \)

Teacher's Note:
1. Smooth contact transfers only translational momentum; rotation is untouched.

Answer: (a) \( 3\pi \) radians

Explanation:
1. The centre of B moves on a circle of radius \( R+2R=3R \): path length \( 6\pi R \).
2. \( \theta=\frac{6\pi R}{2R}=3\pi \) radians.

Teacher's Note:
1. The rotation angle of a rolling body is the distance moved by its centre divided by its radius.

Answer: (c) v

Explanation:
1. \( x=L\cos\theta\Rightarrow v=L\sin\theta\,\dot\theta\Rightarrow\dot\theta=\frac{v}{L\sin\theta} \).
2. The centre moves on a circle of radius L/2 about the wall-floor corner: \( v_c=\frac L2\dot\theta=\frac{v}{2\sin30^\circ}=v \)

Teacher's Note:
1. The centre of a sliding ladder moves along a circle of radius L/2.

Answer: (a) \( \frac{2v^2}{L^2} \)

Explanation:
1. \( \dot\theta=-\frac{v}{L\sin\theta} \); differentiating with constant v: \( |\ddot\theta|=\frac{v\cos\theta}{L\sin^2\theta}|\dot\theta|=\frac{v^2\cos\theta}{L^2\sin^3\theta} \).
2. At 45 degrees \( \frac{\cos\theta}{\sin^3\theta}=2 \), so \( \frac{2v^2}{L^2} \)

Teacher's Note:
1. Differentiate the angular speed expression with respect to time keeping v constant.

Answer: (c) \( \frac{ML^2}{6} \)

Explanation:
1. Height of the plate \( h=L/2 \); centroid is \( \frac{2h}3=\frac L3 \) from O.
2. About the z-axis through the centroid: \( \frac{Mh^2}{18}+\frac{ML^2}{24}=\frac{ML^2}{72}+\frac{ML^2}{24}=\frac{ML^2}{18} \).
3. \( I_O=\frac{ML^2}{18}+M\frac{L^2}9=\frac{ML^2}{6} \)

Teacher's Note:
1. Use the standard triangle results and the parallel axis theorem.

Answer: (a) \( \frac{ML^2}{8} \)

Explanation:
1. About the centroidal axis parallel to the base: \( \frac{Mh^2}{18} \).
2. Shift by \( \frac{2h}3 \): \( I_x=\frac{Mh^2}{18}+\frac{4Mh^2}9=\frac{Mh^2}2=\frac{ML^2}{8} \)

Teacher's Note:
1. The x-axis passes through the apex, a distance 2h/3 from the centroid.

Answer: (c) \( \frac{ML^2}{24} \)

Explanation:
1. For a triangle about its base: \( I=\frac{Mh^2}{6}=\frac{M}{6}\cdot\frac{L^2}4=\frac{ML^2}{24} \)

Teacher's Note:
1. A triangle's moment of inertia about its base is M h squared over 6.

Answer: (b) Down at 20 degrees to the horizontal

Explanation:
1. For a thin rod the angular momentum about the pivot is perpendicular to the rod: \( L=\frac{mL^2}{3}\omega\sin20^\circ \).
2. A line perpendicular to a rod at 20 degrees from the vertical is at 20 degrees from the horizontal. Its component along \( \vec\omega \) (downward for clockwise rotation seen from above) is positive, so it points down at 20 degrees to the horizontal.

Teacher's Note:
1. A thin rod has no angular momentum component along its own length.

Answer: (b) Yes, horizontally

Explanation:
1. The angular momentum has a constant vertical component and a horizontal component that rotates with the rod.
2. So \( \vec\tau=\frac{d\vec L}{dt} \) is non-zero, horizontal and perpendicular to the horizontal part of L.

Teacher's Note:
1. The changing direction of L needs a torque, supplied by the axle at the pivot.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A is true by the parallel axis theorem \( I=I_{cm}+md^2 \). R is also true.
2. But R is about torque of weight, not about the moment of inertia, so it does not explain A.

Teacher's Note:
1. Both true, but unrelated in cause.

Answer: (d) A is false, but R is true.

Explanation:
1. The parallel axis theorem applies only to axes parallel to the one through the centre of mass; two axes at equal distance may have different directions and different \( I_{cm} \).
2. So A is false and R is true.

Teacher's Note:
1. Equal distance does not mean equal I unless the axes are parallel.

Answer: (d) A is false, but R is true.

Explanation:
1. Work done \( =\tau\theta \) is the same for both over one rotation, so the kinetic energies are equal; A is false.
2. R is true (\( \frac12MR^2>\frac25MR^2 \)).

Teacher's Note:
1. Equal torque through equal angle means equal work.

Answer: (c) A is true, but R is false.

Explanation:
1. He leans so that the torque of the friction and normal reaction about the centre of mass balances; the lean is not meant to lower the centre of gravity.
2. A is true and R is false.

Teacher's Note:
1. Leaning provides the required torque balance on the curve.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Since all distances are fixed, the relative motion of any point about another is pure rotation with the common angular velocity.
2. So R explains A.

Teacher's Note:
1. Fixed distances imply a common angular velocity.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. In the CM frame the centre of mass is at rest, so the only motion is rotation about it, and \( K=\frac12I_{cm}\omega^2 \).

Teacher's Note:
1. No translational motion in the CM frame.

Answer: (d) A is false, but R is true.

Explanation:
1. Only the component of L along the axis equals \( I\omega \); for asymmetric bodies L is not parallel to the angular velocity, so A is false.
2. R is true.

Teacher's Note:
1. Angular momentum and angular velocity are parallel only about principal axes.

Answer: (c) \( \frac\omega3 \)

Explanation:
1. The particle's velocity is radial, so it brings no angular momentum about the axis.
2. \( \frac12mR^2\omega=\left(\frac12mR^2+mR^2\right)\omega'\Rightarrow\omega'=\frac\omega3 \)

Teacher's Note:
1. Conserve angular momentum about the axis; a radial approach adds only moment of inertia.

Answer: (b) 0.77 m/s squared

Explanation:
1. \( a=\frac{(m_A-m_B)g}{m_A+m_B+M/2}=\frac{2\times10}{4+2+20}=0.77\ \text{m/s}^2 \)

Teacher's Note:
1. Treat the disc as adding an effective mass M/2 to the system.

Answer: (b) 61

Explanation:
1. \( \alpha=\frac aR=\frac{0.77}{0.1}=7.7\ \text{rad/s}^2 \).
2. \( \theta=\frac12\alpha t^2=\frac12\times7.7\times100=385\ \text{rad}\Rightarrow\frac{385}{2\pi}\approx61\ \text{revolutions} \)

Teacher's Note:
1. Use the angular acceleration from the linear acceleration and a = alpha R.

Answer: (a) \( m\left(\frac{4gh}{\omega^2R^2}-2\right) \)

Explanation:
1. Energy: \( mgh=\frac12mv^2+\frac12\left(\frac12MR^2\right)\omega^2 \) with \( v=\omega R \).
2. \( \frac14M\omega^2R^2=mgh-\frac12m\omega^2R^2\Rightarrow M=m\left(\frac{4gh}{\omega^2R^2}-2\right) \)

Teacher's Note:
1. All the lost potential energy goes to the block and the disk.