A) \[\frac{F}{2mR}\]
B) \[\frac{2\,F}{3mR}\]
C) \[\frac{3\,F}{2mR}\]
D) \[\frac{F}{3mR}\]
Correct Answer: B
Solution :
\[{{_{\tau abt}}_{A\,}}\,=\,\,{{I}_{A\alpha }}\] \[F\times R=\left( \frac{1}{2}M{{R}^{2}}+M{{R}^{2}} \right)\alpha \] \[\alpha \,\,=\,\,\frac{2F}{3mR}\] Note: In this question value of force and its point of application is not given so its answer cannot be given. We are giving its answer by assuming that F is applied at centre of cylinder.You need to login to perform this action.
You will be redirected in
3 sec