System of Particles and Rotational Motion - Result Question 39
41. 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
[2005]
(a) $2: 1$
(b) $1: 2$
(c) $\sqrt{2}: 1$
(d) $1: \sqrt{2}$
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Answer:
Correct Answer: 41. (d)
Solution:
- (d) The relation between K.E. of rotation and angular momentum,
$ \begin{aligned} & K=\frac{L^{2}}{2 I} \Rightarrow L^{2}=2 K I \Rightarrow L=\sqrt{2 K I} \\ & \frac{L_1}{L_2}=\sqrt{(\frac{K_1}{K_2})(\frac{I_1}{I_2})}=\sqrt{(\frac{K}{K})(\frac{I}{2 I})}=\frac{1}{\sqrt{2}} \\ & L_1: L_2=1: \sqrt{2} \end{aligned} $
(d) $I=2 kgm^{2}, v_0=60 rpm=1 rps$
$\omega_0=2 \pi v_0=2 \pi rad / sec$
$\omega_f=0$ and $t=1 min=60 sec$
So, $\propto=\frac{\omega_f-\omega_0}{t}=\frac{0-2 \pi}{60}$
$\alpha=\frac{-\pi}{30} rad / sec^{2}$
and Torque, $\tau=I . \alpha$
$ \begin{aligned} & \tau=2 .(\frac{-\pi}{30})=\frac{-\pi}{15} \\ & \tau=\frac{\pi}{15} N-m \end{aligned} $
