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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