| A uniform rod of length \[l\] and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is: [AIPMT (S) 2006] |
| (Moment of inertia of rod about A is \[\frac{m{{l}^{2}}}{3}\]) |
 |
A) \[\frac{3g}{2l}\]
B) \[\frac{2l}{3g}\]
C) \[\frac{3g}{2{{l}^{2}}}\]
D) \[mg\frac{l}{2}\]
Correct Answer: A
Solution :
| The moment of inertia of the uniform rod about an axis through one end and perpendicular to length is |
| \[I=\frac{m{{l}^{2}}}{3}\] |
| where m is mass of rod and \[l\] its length. |
| Torque \[(\tau =I\alpha )\] acting on centre of gravity of rod is given by |
| \[\tau =mg\frac{l}{2}\] |
| or \[I\alpha =mg\frac{l}{2}\] |
| or \[\frac{m{{l}^{2}}}{3}\alpha =mg\frac{l}{2}\] |
| \[\therefore \] \[\alpha =\frac{3g}{2l}\] |
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