A) 62.5 T-s
B) 70.4 T-s
C) 79.6 T-s
D) 60.5 T-s
Correct Answer: A
Solution :
If I be the moment of inertia of the cylinder about its axis, then \[I=\frac{1}{2}M{{R}^{2}}=\frac{1}{2}\times 20\times {{(0.25)}^{2}}=0.625\,kg-{{m}^{2}}\] \[\therefore \,KE\] associated with the rotating cylinder is given by, \[KE=\frac{1}{2}I{{\omega }^{2}}=\frac{1}{2}\times 0.625\times {{(100)}^{2}}=3125\,J\] Also using the relation, \[L=\sqrt{2I\,(KE)}\] We have \[L=\sqrt{2\times 0.625\times 3125}\] \[L=62.5\,J-s\]
You need to login to perform this action.
You will be redirected in
3 sec
