A) the angular velocity of the rod is \[\frac{3J}{m\ell }\]
B) the angular momentum of the rod is\[J\ell \]
C) the linear velocity of mid point of the rod
D) all the above
Correct Answer: D
Solution :
From angular impulse - momentum theorem \[J\times \ell =I\omega =\frac{m{{\ell }^{2}}}{3}\omega \]\[\Rightarrow \]\[\omega =\frac{3J}{m\ell }\] Angular momentum, \[L=l\omega =J\ell \] Velocity of mid point of rod is \[v=\frac{\ell }{2}\omega =\frac{3J}{2m}\]You need to login to perform this action.
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