JEE Physics System Of Particles and Rotational Motion MCQs Set 58

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

Explanation:
1. The height of the triangle is \( h=\frac L2 \).
2. About an axis through the apex parallel to the base, \( I=\frac{Mh^2}{2}=\frac{ML^2}{8} \).

Teacher's Note:
1. A thin strip at distance \( y \) from the apex has mass proportional to \( y \).

Answer: (d) \( 300 \) rad/s (first, reversed) and \( 150 \) rad/s

Explanation:
1. Let the friction impulse be \( J \) and \( u=\frac JR \). With clockwise positive: \( \omega_1'=100-\frac{JR}{I_1}=100-2u \) and \( \omega_2'=200-\frac{2RJ}{I_2}=200-\frac u4 \) (\( I_1=\frac12mR^2 \), \( I_2=\frac12(4m)(2R)^2 \)).
2. No slipping at the contact: \( \omega_1'R=-\omega_2'(2R) \), i.e. \( \omega_1'=-2\omega_2' \).
3. \( 100-2u=-400+\frac u2 \) gives \( u=200 \), so \( \omega_1'=-300 \) rad/s (reversed) and \( \omega_2'=150 \) rad/s.

Teacher's Note:
1. Touching rotors end up turning in opposite senses once the contact point has no slipping.

Answer: (b) \( \sqrt2\,\frac Jm \)

Explanation:
1. \( v_{cm}=\frac Jm \) and \( \omega=\frac{J(l/2)}{ml^2/12}=\frac{6J}{ml} \).
2. In time \( t \) the rod turns through \( \omega t=\frac\pi2 \).
3. The speed of P relative to the centre is \( \omega\frac l6=\frac Jm \) and it is now perpendicular to \( v_{cm} \), so \( v_P=\sqrt{\left(\frac Jm\right)^2+\left(\frac Jm\right)^2}=\sqrt2\frac Jm \).

Teacher's Note:
1. The centre moves uniformly while P moves in a circle around it.

Answer: (a) \( \frac{3v_0^2}{8\mu g} \)

Explanation:
1. The ring starts rolling at \( t=\frac{v_0}{2\mu g} \) with \( v=\frac{v_0}{2} \).
2. \( s=\frac{v_0^2-v^2}{2\mu g}=\frac{v_0^2-\frac{v_0^2}{4}}{2\mu g}=\frac{3v_0^2}{8\mu g} \).

Teacher's Note:
1. Use \( v^2=v_0^2-2\mu g s \) with the final rolling speed \( \frac{v_0}{2} \).

Answer: (b) \( 5 \) N m

Explanation:
1. \( \tau_{avg}=\frac{\Delta L}{\Delta t}=\frac{10}{2}=5 \) N m (in magnitude).

Teacher's Note:
1. Torque is the rate of change of angular momentum.

Answer: (d) \( \sqrt2 \)

Explanation:
1. Spring energy equals rotational energy: \( \frac12kx_1^2=\frac12I(2\omega)^2=2I\omega^2 \) and \( \frac12kx_2^2=\frac12(2I)\omega^2=I\omega^2 \).
2. \( \frac{x_1^2}{x_2^2}=2 \), so \( \frac{x_1}{x_2}=\sqrt2 \).

Teacher's Note:
1. The stored spring energy is converted entirely into rotational kinetic energy.

Answer: (b) its angular velocity increases but the force on the hinge remains constant

Explanation:
1. Constant torque \( FR \) gives constant angular acceleration, so \( \omega \) increases.
2. The centre does not move, so the hinge force balances \( F \) and the weight; it stays constant.

Teacher's Note:
1. Constant torque means constant angular acceleration, not constant angular velocity.

Answer: (c) \( 75 \) J

Explanation:
1. Angular momentum about the hinge: \( J\ell=10 \) J s, and \( I=\frac{m\ell^2}{3}=\frac23 \) kg m\( ^2 \).
2. \( K=\frac{L^2}{2I}=\frac{100}{2\times\frac23}=75 \) J.

Teacher's Note:
1. Use \( K=\frac{L^2}{2I} \).

Answer: (a) \( 40 \) rad

Explanation:
1. \( mg-T=ma \) and \( TR=\frac12MR^2\frac aR \) give \( T=\frac{Ma}{2}=a \) and \( 10-a=a \), so \( a=5 \) m/s\( ^2 \).
2. \( \alpha=\frac aR=5 \) rad/s\( ^2 \), so \( \theta=\frac12\alpha t^2=\frac12(5)(16)=40 \) rad.

Teacher's Note:
1. Combine the translational equation of the block with the rotational equation of the disc.

Answer: (d) \( 16 \) kg m\( ^2 \)/s

Explanation:
1. Perpendicular distance of the line from the origin: \( \frac{5}{\sqrt{3^2+4^2}}=1 \) m.
2. \( L=mvr_\perp=2\times8\times1=16 \) kg m\( ^2 \)/s.

Teacher's Note:
1. For straight-line motion \( L=mv\times \) (perpendicular distance).

Answer: (d) \( 0.75 \) m

Explanation:
1. \( t=\frac{1}{\sin\theta}\sqrt{\frac{2h}{kg}} \) with \( k=\frac12 \) (ring) and \( k=\frac23 \) (disc), \( \sin60^\circ=\frac{\sqrt3}{2} \).
2. \( t_{ring}-t_{disc}=\frac{2}{\sqrt3}\sqrt{\frac hg}(2-\sqrt3)=\frac{2-\sqrt3}{\sqrt{10}} \), giving \( \sqrt h=\frac{\sqrt3}{2} \), \( h=0.75 \) m.

Teacher's Note:
1. Compare the times using \( a=kg\sin\theta \).

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

Explanation:
1. The height is \( h=\frac L2 \). About the base, \( I=\frac{Mh^2}{6} \).
2. \( I=\frac M6\cdot\frac{L^2}{4}=\frac{ML^2}{24} \).

Teacher's Note:
1. For a triangle, \( I_{base}=\frac{Mh^2}{6} \).

Answer: (a) \( \frac{10}{3} \) m/s

Explanation:
1. The string jerks and angular momentum about the hinge is conserved: \( mv_0r=mvr+\frac12mr^2\omega \) with \( v=\omega r \).
2. \( v_0=v+\frac v2 \), so \( v=\frac23v_0=\frac{10}{3} \) m/s.

Teacher's Note:
1. The hinge force passes through the axis, so angular momentum about the hinge is conserved during the jerk.

Answer: (b) \( \frac83mvR \)

Explanation:
1. Translation: total mass \( 2m \) with speed \( v \) at height \( R \): \( L_{trans}=2mvR \).
2. Spin of the shell only: \( I\omega=\frac23mR^2\cdot\frac vR=\frac23mvR \).
3. \( L=2mvR+\frac23mvR=\frac83mvR \).

Teacher's Note:
1. Angular momentum about a ground point is \( L_{cm} \) (orbital) plus spin.

Answer: (b) \( \frac16 \) m

Explanation:
1. The block is on the verge of slipping when \( \tan\theta=\mu \), with the slope \( \frac{dy}{dx}=\frac{x^2}{2} \).
2. \( \frac{x^2}{2}=0.5 \) gives \( x=1 \) m, so \( y=\frac{1}{6} \) m.

Teacher's Note:
1. Slipping starts when the slope equals the coefficient of friction.

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

Explanation:
1. \( M=\int_0^LKx\,dx=\frac{KL^2}{2} \).
2. \( I=\int_0^LKx\cdot x^2\,dx=\frac{KL^4}{4}=\frac{ML^2}{2} \).

Teacher's Note:
1. Integrate \( \lambda\,x^2\,dx \) over the length.

Answer: (c) \( \frac{n_2mx^2}{n_1-n_2} \)

Explanation:
1. Angular momentum conservation: \( In_1=(I+mx^2)n_2 \).
2. \( I(n_1-n_2)=mx^2n_2 \), so \( I=\frac{n_2mx^2}{n_1-n_2} \).

Teacher's Note:
1. Sticking the mass raises the moment of inertia and lowers the rotation rate.

Answer: (c) \( \frac12Mr^2 \)

Explanation:
1. Two such semicircular discs make a full disc of mass \( 2M \) with \( I=\frac12(2M)r^2=Mr^2 \).
2. By symmetry each half has half of that: \( \frac12Mr^2 \).

Teacher's Note:
1. The half disc has the same \( \frac12MR^2 \) formula as a full disc about the centre.

Answer: (c) \( \frac{5a}{9} \)

Explanation:
1. \( M=\rho_0\int_0^a\left(1+\frac xa\right)dx=\frac32\rho_0a \).
2. \( x_{cm}=\frac{\rho_0\int_0^ax\left(1+\frac xa\right)dx}{M}=\frac{\rho_0\left(\frac{a^2}{2}+\frac{a^2}{3}\right)}{\frac32\rho_0a}=\frac{5a}{9} \).

Teacher's Note:
1. The denser end (B) pulls the centre of mass beyond the midpoint.

Answer: (c) \( 2m\left(\frac{2gh}{\omega^2R^2}-1\right) \)

Explanation:
1. Energy: \( mgh=\frac12m(\omega R)^2+\frac12\cdot\frac12M R^2\omega^2 \).
2. \( \frac14M\omega^2R^2=mgh-\frac12m\omega^2R^2 \), so \( M=\frac{4mgh}{\omega^2R^2}-2m=2m\left(\frac{2gh}{\omega^2R^2}-1\right) \).

Teacher's Note:
1. All the potential energy lost goes into the block's and the disc's kinetic energy.

Answer: (a) \( \sqrt{\frac{6gh}{5}} \)

Explanation:
1. \( mgh=\frac12mv^2+\frac12\cdot\frac23mR^2\frac{v^2}{R^2}=\frac56mv^2 \).
2. \( v=\sqrt{\frac{6gh}{5}} \).

Teacher's Note:
1. Use \( K=\frac12mv^2\left(1+\frac{k^2}{R^2}\right) \) with \( k^2=\frac23R^2 \).

Answer: (c) directly proportional to \( t \)

Explanation:
1. \( \alpha=\frac\tau I \) is constant, so \( \omega=\alpha t \).
2. \( P=\tau\omega=\frac{\tau^2}{I}t \), which is linear in \( t \).

Teacher's Note:
1. Power equals torque times angular velocity.

Answer: (b) \( \frac{7\rho_0a^3}{12} \)

Explanation:
1. \( I=\rho_0\int_0^a\left(1+\frac xa\right)x^2\,dx=\rho_0\left(\frac{a^3}{3}+\frac{a^3}{4}\right)=\frac{7\rho_0a^3}{12} \).

Teacher's Note:
1. Integrate \( x^2\,dm \) with \( dm=\rho_0\left(1+\frac xa\right)dx \).

Answer: (b) horizontal and intersecting the axis

Explanation:
1. Uniform circular motion needs only a centripetal force.
2. That force is horizontal and directed towards the axis, so its line of action meets the axis.

Teacher's Note:
1. There is no tangential acceleration in uniform rotation.

Answer: (d) \( -F(\hat i+\hat j) \)

Explanation:
1. Position of the point of application relative to the point \( (1,-1,0) \): \( \vec r=(-1,1,0) \).
2. \( \vec\tau=\vec r\times\vec F=(-\hat i+\hat j)\times(-F\hat k)=-F\hat i-F\hat j \).

Teacher's Note:
1. Use \( \hat i\times\hat k=-\hat j \) and \( \hat j\times\hat k=\hat i \).

Answer: (a) \( \frac I2 \)

Explanation:
1. By symmetry the moments of inertia about all in-plane axes through O are equal, say \( I' \).
2. Perpendicular axis theorem: \( I=I'+I' \), so \( I'=\frac I2 \).

Teacher's Note:
1. For a square, every in-plane axis through the centre is a principal axis with equal moment.