A heavy rod of weight W is held in a horizont | vedclass

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(D) Let the length of the rod be $l$ and its mass be $m$,so $W = mg$. Initially,the rod is in equilibrium. When one man at end $B$ lets go,the rod sta

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$W/4$

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(D) Let the length of the rod be $l$ and its mass be $m$,so $W = mg$. Initially,the rod is in equilibrium. When one man at end $B$ lets go,the rod starts rotating about the other end $A$ due to the torque produced by its weight acting at the center of mass $G$ (at distance $l/2$ from $A$).The torque $	au$ about point $A$ is given by $	au = W \cdot \frac{l}{2}$.The moment of inertia $I$ of the rod about point $A$ is $I = \frac{ml^2}{3}$.Using $	au = I\alpha$,we have $W \cdot \frac{l}{2} = \left(\frac{ml^2}{3}ight)\alpha$.Substituting $W = mg$,we get $mg \cdot \frac{l}{2} = \frac{ml^2}{3} \alpha$,which simplifies to $\alpha = \frac{3g}{2l}$.The linear acceleration $a$ of the center of mass $G$ is $a = \frac{l}{2} \alpha = \frac{l}{2} \cdot \frac{3g}{2l} = \frac{3g}{4}$.Now,applying Newton's second law for the vertical motion of the center of mass: $W - N = ma$,where $N$ is the normal force exerted by the man at $A$.$mg - N = m \left(\frac{3g}{4}ight)$.$N = mg - \frac{3mg}{4} = \frac{mg}{4} = \frac{W}{4}$.Thus,the other man feels a force of $W/4$.

$(a)$ $h = \frac{R}{2}$	$(i)$ Sphere rolls without slipping with a constant velocity and no loss of energy.
$(b)$ $h = R$	$(ii)$ Sphere spins clockwise,loses energy by friction.
$(c)$ $h = \frac{3R}{2}$	$(iii)$ Sphere spins anti-clockwise,loses energy by friction.
$(d)$ $h = \frac{7R}{5}$	$(iv)$ Sphere has only a translational motion,loses energy by friction.

$A$ heavy rod of weight $W$ is held in a horizontal position by two men at its ends. If one man suddenly lets go,what force will the other man feel?

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