A) \[T=\sqrt{\frac{2\pi m}{IB}}\]
B) \[T=\sqrt{\frac{\pi m}{IB}}\]
C) \[T=\sqrt{\frac{2m}{IB}}\]
D) \[T=\sqrt{\frac{\pi m}{2IB}}\]
Correct Answer: A
Solution :
\[\tau =MB\sin \theta =I\alpha \] \[\pi {{R}^{2}}IB\theta =\frac{M{{R}^{2}}}{2}\alpha \] \[\omega =\sqrt{\frac{2\pi IB}{m}}\] \[T=\sqrt{\frac{2\pi M}{IB}}\]
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