JEE Physics System Of Particles and Rotational Motion MCQs Set 20

Answer: (d) \( \frac{3\rho L^3}{8\pi^2} \)

Explanation:
1. Mass \( M=\rho L \) and radius \( r=\frac{L}{2\pi} \).
2. About a tangent, by the parallel axis theorem: \( I=\frac12Mr^2+Mr^2=\frac32Mr^2 \).
3. \( I=\frac32\rho L\cdot\frac{L^2}{4\pi^2}=\frac{3\rho L^3}{8\pi^2} \)

Teacher's Note:
1. A ring has I = (1/2)Mr² about a diameter and Mr² about its axis.
2. The tangent is parallel to a diameter at distance r, so add Mr².

Answer: (b) \( \frac{\omega M}{M+2m} \)

Explanation:
1. No external torque acts, so angular momentum is conserved.
2. \( Mr^2\omega=(M+2m)r^2\omega' \).
3. \( \omega'=\frac{M\omega}{M+2m} \)

Teacher's Note:
1. Attaching masses gently on the rim adds moment of inertia 2mr² without any external torque.
2. More inertia at the same angular momentum means a smaller angular velocity.

Answer: (a) 0.4 cm

Explanation:
1. Masses are proportional to areas: removed disc \( \propto 2^2=4 \), original \( \propto6^2=36 \), remainder \( \propto32 \).
2. Shift of the centre of mass: \( \frac{m_{removed}\,d}{m_{remaining}}=\frac{4\times3.2}{32}=0.4\ \text{cm} \).

Teacher's Note:
1. Treat the removed part as a negative mass at its own centre.
2. The shift is directed away from the removed portion.

Answer: (b) \( \sqrt{30} \)

Explanation:
1. Fall of the centre of mass: \( h=\frac L2\sin30^\circ=0.25\times0.5=0.125\ \text{m} \).
2. Energy conservation with \( I=\frac{ML^2}{3} \): \( \frac12\cdot\frac{ML^2}{3}\omega^2=Mgh \).
3. \( \omega^2=\frac{6gh}{L^2}=\frac{6\times10\times0.125}{0.25}=30 \), so \( \omega=\sqrt{30}\ \text{rad/s} \).

Teacher's Note:
1. Use potential energy lost by the centre of mass and rotational kinetic energy about the pivot.
2. I about the end of a rod is ML²/3.

Answer: (b) 16 rad/s²

Explanation:
1. Take torques about the contact point: the force acts at height 2R, so \( \tau=F\cdot2R=40\times1=40\ \text{N m} \).
2. Hollow cylinder about contact point: \( I=MR^2+MR^2=2MR^2=2.5\ \text{kg m}^2 \).
3. \( \alpha=\frac{40}{2.5}=16\ \text{rad/s}^2 \)

Teacher's Note:
1. For pure rolling, torque about the instantaneous contact point equals I_contact × α.
2. The force applied at the top of the cylinder has lever arm 2R about the contact point.

Answer: (a) \( \vec r_{cm}=\frac{13}{8}L\hat x+\frac58L\hat y \)

Explanation:
1. Three straight pieces with masses proportional to length: top piece 2L with centre \( (L,\,L) \), vertical piece L with centre \( (2L,\,L/2) \), bottom piece L with centre \( (2.5L,\,0) \).
2. Total length 4L.
3. \( x_{cm}=\frac{2L\cdot L+L\cdot2L+L\cdot2.5L}{4L}=\frac{13}{8}L \).
4. \( y_{cm}=\frac{2L\cdot L+L\cdot\frac L2+0}{4L}=\frac58L \).

Teacher's Note:
1. For a uniform wire, mass is proportional to length, and each straight piece acts at its midpoint.
2. Compute x and y separately.

Answer: (d) \( \left(\frac{5a}{12},\frac{5b}{12}\right) \)

Explanation:
1. Treat the full sheet (mass 4m, centre \( (\frac a2,\frac b2) \)) as remaining part plus removed quarter (mass m, centre \( (\frac{3a}{4},\frac{3b}{4}) \)).
2. \( x_{cm}=\frac{4m\cdot\frac a2-m\cdot\frac{3a}{4}}{3m}=\frac{5a}{12} \), and similarly \( y_{cm}=\frac{5b}{12} \).

Teacher's Note:
1. Removing mass is like adding a negative mass at the centre of the removed part.
2. The centre of mass shifts away from the cut-out corner.

Answer: (a) 2 s

Explanation:
1. \( \frac12I\omega^2=1200\Rightarrow\omega^2=\frac{2400}{1.5}=1600\Rightarrow\omega=40\ \text{rad/s} \).
2. \( \omega=\alpha t\Rightarrow t=\frac{40}{20}=2\ \text{s} \)

Teacher's Note:
1. First find the angular speed needed from the kinetic energy.
2. Then use ω = αt starting from rest.

Answer: (a) 3 J

Explanation:
1. Total kinetic energy of a rolling disc: \( K=\frac12mv^2\left(1+\frac12\right)=\frac34mv^2 \).
2. \( K=\frac34\times100\times(0.2)^2=3\ \text{J} \).
3. Work needed to stop it equals this kinetic energy, 3 J.

Teacher's Note:
1. A rolling body has translational plus rotational kinetic energy.
2. Convert 20 cm/s to 0.2 m/s before substituting.

Answer: (c) \( \frac85ma^2+2mb^2 \)

Explanation:
1. Two spheres have their centres on the axis: each contributes \( \frac25ma^2 \).
2. The other two are at distance b: each contributes \( \frac25ma^2+mb^2 \).
3. Total: \( 4\times\frac25ma^2+2mb^2=\frac85ma^2+2mb^2 \)

Teacher's Note:
1. Use I = (2/5)ma² about a diameter and add mb² for spheres displaced by b.
2. Count the spheres on the axis and off it separately.

Answer: (b) \( \frac23MR^2 \)

Explanation:
1. Energy conservation: \( \frac12Mv^2+\frac12I\frac{v^2}{R^2}=Mgh=\frac{5Mv^2}{6} \).
2. \( \frac12+\frac{I}{2MR^2}=\frac56\Rightarrow\frac{I}{2MR^2}=\frac13 \), so \( I=\frac23MR^2 \)

Teacher's Note:
1. All the initial kinetic energy (translational + rotational) becomes potential energy at the top.
2. Replace ω by v/R for rolling without slipping.

Answer: (a) 16 m/s

Explanation:
1. Initially only the heavier block moves, so \( v_{cm}=\frac{20\times20+5\times0}{25}=16\ \text{m/s} \).
2. No external horizontal force acts afterwards, so the velocity of the centre of mass stays 16 m/s.

Teacher's Note:
1. The spring force is internal, so it cannot change the centre-of-mass velocity.
2. Compute v_cm right after the impulse.

Answer: (b) \( \frac{m_2L}{m_1+m_2} \)

Explanation:
1. Place m1 at the origin and m2 at x = L.
2. \( x_{cm}=\frac{m_1\cdot0+m_2L}{m_1+m_2}=\frac{m_2L}{m_1+m_2} \)

Teacher's Note:
1. The centre of mass lies closer to the heavier particle.
2. If m1 is very large, x_cm tends to 0, which agrees with the formula.

Answer: (c) \( \frac54ma^2 \)

Explanation:
1. The mass at O lies on the axis: contribution 0.
2. A is at distance a from the axis (OA is perpendicular to it): \( ma^2 \).
3. B is at horizontal distance \( \frac a2 \) from the axis: \( m\frac{a^2}{4} \).
4. Total: \( ma^2+\frac{ma^2}{4}=\frac54ma^2 \)

Teacher's Note:
1. The distance of a point from the axis is its perpendicular distance, here the horizontal coordinate.
2. B lies above the midpoint of OA, so its horizontal distance is a/2.

Answer: (a) \( v_Q>v_C>v_P \)

Explanation:
1. For rolling, a point at height y above the contact point has speed \( \omega y \).
2. Q: \( \omega\cdot\frac{3R}{2} \); C: \( \omega R \); P: \( \omega\cdot\frac R2 \).
3. Hence \( v_Q>v_C>v_P \)

Teacher's Note:
1. Rolling is instantaneous rotation about the contact point, so speed grows with distance from the contact point.
2. The top of the disc moves at 2v_C and the bottom is at rest.

Answer: (b) inside the square plate

Explanation:
1. Masses are proportional to areas: circle \( \frac{\pi d^2}{4}\approx0.785d^2 \), square \( d^2 \).
2. Take the circle's centre as origin; the square's centre is at \( x=d \).
3. \( x_{cm}=\frac{0.785d^2\cdot0+d^2\cdot d}{1.785d^2}\approx0.56d \), while the contact point is at \( x=0.5d \).
4. So the centre of mass lies beyond the contact point, inside the square plate (which spans 0.5d to 1.5d).

Teacher's Note:
1. The centre of mass lies on the line joining the two centres, closer to the heavier part.
2. Compare x_cm with the position of the contact point to decide which plate contains it.

Answer: (d) R/6

Explanation:
1. Masses are proportional to areas: hole \( \propto\left(\frac R2\right)^2=\frac{R^2}{4} \), original \( \propto R^2 \), so the remaining mass is \( \frac34 \) of the original and the hole is \( \frac14 \).
2. \( x_{cm}=\frac{M\cdot0-\frac M4\cdot\frac R2}{\frac{3M}{4}}=-\frac R6 \).
3. The centre of gravity is R/6 from O, on the side opposite to the hole.

Teacher's Note:
1. Treat the hole as a negative mass placed at its own centre.
2. Hole mass : remaining mass = 1 : 3.

Answer: (d) \( \frac72MR^2 \)

Explanation:
1. Top ring: the axis is along a diameter, so \( I_1=\frac12MR^2 \).
2. Each lower ring: its centre is at distance R from the axis, so \( I=\frac12MR^2+MR^2=\frac32MR^2 \).
3. Total: \( \frac12MR^2+2\times\frac32MR^2=\frac72MR^2 \)

Teacher's Note:
1. A ring has I = MR²/2 about any of its diameters.
2. Use the parallel axis theorem for the two lower rings, whose centres are R away from YY'.

Answer: (a) \( \sqrt{\frac{10}{7}gh} \)

Explanation:
1. \( mgh=\frac12mv^2+\frac12\cdot\frac25mR^2\cdot\frac{v^2}{R^2}=\frac7{10}mv^2 \).
2. \( v=\sqrt{\frac{10}{7}gh} \)

Teacher's Note:
1. Potential energy is shared between translation and rotation.
2. General result: v = √(2gh/(1 + k)).

Answer: (b) \( \frac{3g}{2\ell} \)

Explanation:
1. Torque of the weight about A: \( \tau=mg\frac\ell2 \).
2. \( \alpha=\frac\tau I=\frac{mg\ell/2}{m\ell^2/3}=\frac{3g}{2\ell} \)

Teacher's Note:
1. The weight acts at the centre of the rod, a distance ℓ/2 from the pivot.
2. Check units: g/ℓ has units of 1/s², which is the unit of angular acceleration.

Answer: (b) \( \frac{MR^2}{4} \)

Explanation:
1. The X-axis passes through the centre of the disc and lies in its plane, so it is a diameter.
2. For a disc, \( I_{diameter}=\frac14MR^2 \). The distance a plays no role since the axis passes through the centre.

Teacher's Note:
1. Moment of inertia about a diameter is half of that about the perpendicular axis through the centre.
2. Do not add Ma² unless the axis is displaced from the centre.

Answer: (b) \( \frac{ML^2}{9} \)

Explanation:
1. The axis divides the rod into pieces of length L/3 (mass M/3) and 2L/3 (mass 2M/3), each with the axis at one end.
2. \( I=\frac13\cdot\frac M3\cdot\left(\frac L3\right)^2+\frac13\cdot\frac{2M}{3}\cdot\left(\frac{2L}{3}\right)^2=\frac{ML^2}{81}+\frac{8ML^2}{81}=\frac{ML^2}{9} \).
3. Check with the parallel axis theorem: \( \frac{ML^2}{12}+M\left(\frac L2-\frac L3\right)^2=\frac{ML^2}{12}+\frac{ML^2}{36}=\frac{ML^2}{9} \)

Teacher's Note:
1. Both methods (splitting the rod, or the parallel axis theorem with d = L/6) must agree.
2. I is smallest at the centre (ML²/12) and increases as the axis moves outward.

Answer: (a) L

Explanation:
1. \( \tau=\frac{\Delta L_{ang}}{\Delta t}=\frac{5L-2L}{3}=L \)

Teacher's Note:
1. Torque is the rate of change of angular momentum.
2. Here L is just a symbol for a unit of angular momentum.