1. ⇒ (MHT CET 2023 13th May Morning Shift )

A rigid body rotates with an angular momentum L. If its rotational kinetic energy is made four times, its angular momentum will become

A. 4 L

B. 16 L

C. $\sqrt{2}$ L

D. 2 L

Correct Option is (D)

Angular momentum,

$\begin{aligned} L & = \sqrt{2 KI} \\ K & = 4 K (given) \\ ∴ L & = \sqrt{2 \times 4 KI} = 2 \sqrt{2 KI} = 2 L \end{aligned}$

2. ⇒ (MHT CET 2023 12th May Morning Shift )

A thin uniform circular disc of mass ' $M$ ' and radius ' $R$ ' is rotating with angular velocity ' $ω$ ', in a horizontal plane about an axis passing through its centre and perpendicular to its plane. Another disc of same radius but of mass $(\frac{M}{2})$ is placed gently on the first disc co-axially. The new angular velocity will be

A. $\frac{2}{3} ω$

B. $\frac{4}{5} ω$

C. $\frac{5}{4} ω$

D. $\frac{3}{2} ω$

Correct Option is (A)

Angular momentum $= I ω$

By conservation of angular momentum, $I_{1} ω_{1} = I_{2} ω_{2}$

Here, $I_{1} = \frac{M R^{2}}{2}, I_{2} = \frac{(M + M / 2)}{2} R^{2} = \frac{3 M R^{2}}{4}$

$\begin{aligned} ∴ \frac{{MR}^{2}}{2} ω_{1} = \frac{3 {MR}^{2}}{4} ω_{2} \\ ∴ ω_{2} = \frac{2}{3} ω_{1} \end{aligned}$

3. ⇒ (MHT CET 2023 11th May Evening Shift )

Two bodies have their moments of inertia I and 2I respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momenta will be in the ratio

A. $1 : 2$

B. $\sqrt{2} : 1$

C. $2 : 1$

D. $1 : \sqrt{2}$

Correct Option is (D)

The equation for angular momentum is

$L = \sqrt{2 K_{Rot} \times I}$

So, $L \propto \sqrt{I}$

$∴$ The ratio of angular momentum of the two bodies is

$\frac{L_{1}}{L_{2}} = \sqrt{\frac{I_{1}}{I_{2}}}$

$\frac{L_{1}}{L_{2}} = \sqrt{\frac{I}{2 I}} . . . . (given I_{2} = 2 I)$

$∴ \frac{L_{1}}{L_{2}} = \frac{1}{\sqrt{2}}$

4. ⇒ (MHT CET 2023 11th May Morning Shift )

A particle moves along a circular path with decreasing speed. Hence

A. its resultant acceleration is towards the centre.

B. it moves in a spiral path with decreasing radius.

C. the direction of angular momentum remains constant.

D. its angular momentum remains constant

Correct Option is (C)

The direction of angular moment remains constant as the angular momentum is a vector quantity, and its direction is perpendicular to the plane of motion. As the speed decreases, the linear momentum decreases, but the angular momentum remains constant due to the conservation of angular momentum.

5. ⇒ (MHT CET 2023 11th May Morning Shift )

A disc has mass $M$ and radius $R$ . How much tangential force should be applied to the rim of the disc, so as to rotate with angular velocity ' $ω$ ' in time $t$ ?

A. $\frac{M R ω}{4 t}$

B. $\frac{MR ω}{2 t}$

C. $\frac{MR ω}{t}$

D. $MR ω t$

Correct Option is (B)

Torque: $τ = I α = \frac{{MR}^{2}}{2} \times \frac{ω}{t}$

$\begin{array}{ll} ∴ & τ = \frac{{MR}^{2} ω}{2 t} \\ But τ = R \times F \\ ∴ & F = \frac{τ}{R} = \frac{MR ω}{2 t} \end{array}$

⇒Rotational Dynamics