Rotational Motion Question 180
Two identical discs of mass m and radius r are arranged. If $\alpha $ is the angular acceleration of the lower disc and $a _{cm}$ is acceleration of centre of mass of the lower disc, then relation between
Options:
A) $a _{cm}=\alpha /r$
B) $a _{cm}=2\alpha r$
C) $a _{cm}=\alpha r$
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b]
$ T r=\frac{m r^2}{2} \alpha \text { ..(1) }$
$ \operatorname{Tr}=\frac{m r^2}{2} \alpha \text { …(2) } $
$ \alpha _1=\alpha \text { \quad (3) } $
Acceleration of point $\mathrm{b}$ equals acceleration of point $\mathrm{a}$
$ r \alpha_1=a _{c m}-r \alpha \text { Hence, } 2 r \alpha=a _{c m} \text { ….(4) } $
BETA
