Rotational Motion Question 186
Question: Linear acceleration of hallow cylinder of mass $m _{2}$ is $a _{2}$ Then angular acceleration${{\alpha } _{2}}$ is (given that there is no slipping).
Options:
A) $\frac{a _{2}}{R}$
B) $\frac{(a _{2}+g)}{R}$
C) $\frac{2(a _{2}+g)}{R}$
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c]
$ m_2 g-T=m_2 a_2 .. \text { (i) } $
$ T R=\frac{m_2 R^2}{2} \alpha_2 .. \text { (ii) } $
$ \alpha_1 R=a_2-\alpha_2 R .. \text { (iii) } $
$ T R=(\frac{m_1 R^2}{2}) \alpha_1 .. \text { (iv) } $
$ \alpha_2=\frac{2 T}{m_2 R}=\frac{2}{m_2 R} \cdot(m_2 a_2+m_2 g) $
$ =2(\frac{(a_2+g)}{R}) $
