Rotational Motion Question 2
Question: Initial angular velocity of a circular disc of mass M is C $ {\omega _{1}} $ . Then, two small spheres of mass m are attached gently to two diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc?
[AIEEE 2002]
Options:
A) $ ( \frac{M+m}{M} ){\omega_1} $
B) $ ( \frac{M+m}{m} ){\omega_1} $
C) $ ( \frac{M}{M+4m} ){\omega_1} $
D) $ ( \frac{M}{M+2m} ){\omega_1} $
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] Conservation of angular momentum gives $ \frac{1}{2}MR^{2}{\omega_1}=( \frac{1}{2}MR^{2}+2MR^{2} ){\omega_2} $ $ (\because l _1{\omega_1}=l _2{\omega_2} $ and $ l _1=\frac{1}{2}mR^{2} $ s $ l _2=\frac{1}{2}mR^{2}+2mR^{2}) $
$ \Rightarrow \frac{1}{2}MR^{2}{\omega_1}=\frac{1}{2}R^{2}(M+4m){\omega_2} $
$ \therefore $ $ {\omega_2}=( \frac{M}{M+4m} ){\omega_1} $
