A) 2 \[\omega \alpha t\]
B) \[\omega \alpha t\]
C) \[\frac{3}{2}\,\omega \alpha t\]
D) \[\frac{\omega \alpha t}{2}\]
Correct Answer: A
Solution :
Since, the moment of inertia of thin rod about an axis passing through its centre of mass and perpendicular to its geometrical axis is \[{{I}_{CM}}=\frac{M{{L}^{2}}}{12}\] M.I. of rod about an axis perpendicular to its geometrical axis and passes through one of the ends \[{{I}_{1}}={{I}_{CM}}+M{{(L/2)}^{2}}\] Theorem of parallel axis \[{{I}_{1}}=\frac{M{{L}^{2}}}{12}+\frac{M{{L}^{2}}}{4}\] \[{{I}_{1}}=\frac{M{{L}^{2}}}{3}\]You need to login to perform this action.
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