| A uniform rod AB of length \[l\] and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is \[\frac{m{{l}^{2}}}{3},\] the initial angular acceleration of the rod will be: [AIPMT (S) 2007] |
 |
A) \[\frac{2g}{3l}\]
B) \[mg\frac{l}{2}\]
C) \[\frac{3}{2}gl\]
D) \[\frac{3g}{2l}\]
Correct Answer: C
Solution :
| The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is |
| \[I=\frac{m{{l}^{2}}}{3}\] |
| where m is mass of rod and \[l\]is 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 \[\frac{m{{l}^{2}}}{3}\alpha =mg\frac{l}{2}\] |
| or \[\alpha =\frac{3g}{2l}\] |
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