A rod of length \[\ell \] is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest in a vertical position shown in Figure. Find velocity of the C.M. of the rod, when rod is inclined at an angle \[\theta \] from the vertical. |
A) \[\sqrt{\frac{gl(1-\cos \theta )}{4}}\]
B) \[\sqrt{\frac{gl(1-\sin \theta )}{4}}\]
C) \[\sqrt{\frac{3gl(1-\sin \theta )}{4}}\]
D) \[\sqrt{\frac{3gl(1-\cos \theta )}{4}}\]
Correct Answer: D
Solution :
The fall in position of C.M of the rod, \[h=\frac{\ell }{2}(1-\cos \theta )\] |
In the process, decrease in P.E. is equal to the increase in rotational K.E. of the rod, so \[mgh=\frac{1}{2}I{{\omega }^{2}}\] |
Or \[mg\frac{\ell }{2}(1-\cos \theta )=\frac{1}{2}\left( \frac{m{{\ell }^{2}}}{3} \right) & & & & & & & _{{}}^{{{\omega }^{2}}}\] |
\[\therefore \omega =\sqrt{\frac{3g}{\ell }(1-\cos \theta )}\] |
The velocity of C.M. of the rod \[V{{ & }_{cm}}=\omega r\] |
\[=\sqrt{\frac{3g}{\ell }(1-\cos \theta )}\times \frac{\ell }{2}=\sqrt{\frac{3g\ell (1-cos\theta )}{4}}\]. |
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