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A device (figure) consists of a smooth L-shaped rod located in a horizontal plane and a sleeve A of mass m attached by a weight less spring to a point B. The spring stiffness is equal to x. The whole system rotates with a constant angular velocity omega about a vertical axis passing through the point O. Find the elongation of the spring. How is the result affected by the rotation direction? |
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Answer» SOLUTION :At FIRST draw the free body diagram of the device as, SHOWN. The forces, acting on the sleeve are it's weight, acting vertically downward, spring forces, along the length of the spring and normal reaction by the rod, perpendicular to its length. Let F be the spring force, and `Deltal` be the elongation. From, `F_n=mw_n`: `Nsin theta+Fcos theta=momega^2r` (1) where `r cos theta=(l_0+Deltal)`. Similarly from `F_t=mw_t` `Ncos theta-Fsin theta=0` or, `N=Fsin theta//cos theta` (2) From (1) and (2) `F(sin theta//costheta)*sin theta+F cos theta=momega^2r` `=momega^2(l_0+Deltal)//cos theta` On putting `F=kappa Deltal`, `kappa DELTA l sin^2 theta+kappa Deltal cos^2 theta=momega^2(l_0+Deltal)` on solving, we get, `Deltal=momega^2(l_0)/(kappa-momega^2)=(l_0)/((kappa//momega^2-1))` and it is independent of the DIRECTION of rotation. 
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