A) \[\frac{6F}{MR}\]
B) \[\frac{3F}{MR}\]
C) \[\frac{6F}{5MR}\]
D) \[\frac{4F}{MR}\]
Correct Answer: A
Solution :
If\[\alpha \]be the angular acceleration of the disc, then, using\[{{F}_{net}}=I\alpha ,\] we have, \[\alpha =\frac{{{F}_{net}}}{I}\] \[=\frac{3FR}{(M{{R}^{2}}/2)}=\frac{6F}{MR}\]You need to login to perform this action.
You will be redirected in
3 sec