SATHEE: Rotational Motion Question 67

Rotational Motion Question 67

Question: A circular disk of moment of inertia $I_{t}$ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed $ω_{t}$ . Another disk of moment of inertia $I_{b}$ is dropped coaxially onto the rotating disk.Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed $ω_{f}$ . The energy lost by the initially rotating disc due to friction is

[AIPMT (S) 2010]

Options:

A) $\frac{1}{2} \frac{I_{b}^{2}}{(I_{t} + I_{b})} ω_{i}^{2}$

B) $\frac{1}{2} \frac{I_{t}^{2}}{(I_{t} + I_{b})} ω_{i}^{2}$

C) $\frac{1}{2} \frac{I_{b} - I_{t}}{(I_{t} + I_{b})} ω_{i}^{2}$

D) $\frac{1}{2} \frac{I_{b} I_{t}}{(I_{t} + I_{b})} ω_{i}^{2}$

Show Answer

Answer:

Correct Answer: D

Solution:

Loss of energy,

$Δ E = \frac{1}{2} I_{t} ω_{i}^{2} - \frac{1}{2} \frac{I_{t}^{2} ω_{i}^{2}}{2 (I_{t} + I_{b})}$ $= \frac{1}{2} \frac{I_{b}^{} I_{t} ω_{i}^{2}}{(I_{t} + I_{b})}$

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Rotational Motion Question 67

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