A rod hinged at one end is released from the | vedclass

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(C) Let the length of the rod be $l$ and its mass be $m$. When the rod is released from the horizontal position,by the conservation of mechanical ener

Vedclass · Accepted Answer

$60$

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(C) Let the length of the rod be $l$ and its mass be $m$. When the rod is released from the horizontal position,by the conservation of mechanical energy,the potential energy lost equals the rotational kinetic energy gained when it reaches the vertical position:$mg(l/2) = (1/2)I\omega^2$Since $I = ml^2/3$ for a rod hinged at one end,we have:$mgl/2 = (1/2)(ml^2/3)\omega^2$$\omega^2 = 3g/l$When the rod reaches the vertical position,the lower half (mass $m/2$) separates. The upper half (mass $m/2$) remains hinged at the top and continues to rotate with the same angular velocity $\omega = \sqrt{3g/l}$.Now,applying the work-energy theorem to the upper half as it swings up to an angle $	heta$ with the vertical:The change in kinetic energy is $\Delta K = K_f - K_i = 0 - (1/2)I_{upper}\omega^2$,where $I_{upper} = (m/2)(l/2)^2/3 = ml^2/24$.The work done by gravity is $W_g = -(m/2)g(h_{cm,f} - h_{cm,i}) = -(m/2)g((l/4) - (l/4)\cos	heta) = -(mgl/8)(1 - \cos	heta)$.Equating $W_g = \Delta K$:$-(mgl/8)(1 - \cos	heta) = -(1/2)(ml^2/24)(3g/l)$$(mgl/8)(1 - \cos	heta) = (mgl/16)$$1 - \cos	heta = 1/2$$\cos	heta = 1/2$$	heta = 60^o$.

$A$ rod hinged at one end is released from the horizontal position as shown in the figure. When it becomes vertical,its lower half separates without exerting any reaction at the breaking point. Then the maximum angle '$\theta$' made by the hinged upper half with the vertical is ......... $^o$.

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