A thin uniform stick of length l and mass m i | vedclass

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Question

(A) The torque $	au$ about the hinge point $B$ due to gravity acting at the centre of mass (at distance $l/2$ from $B$) is given by $	au = mg 	imes

Vedclass · Accepted Answer

$\frac{3}{4} g$

Vedclass · Answer

(A) The torque $	au$ about the hinge point $B$ due to gravity acting at the centre of mass (at distance $l/2$ from $B$) is given by $	au = mg 	imes \frac{l}{2}$.Using the rotational form of Newton's second law,$	au = I\alpha$,where $I$ is the moment of inertia of the stick about the hinge $B$ $(I = \frac{ml^2}{3})$.Thus,$\alpha = \frac{	au}{I} = \frac{mg(l/2)}{ml^2/3} = \frac{3g}{2l}$.The linear acceleration $a$ of the centre of mass is given by $a = \alpha r$,where $r = l/2$ is the distance of the centre of mass from the hinge.Therefore,$a = \left(\frac{3g}{2l}ight) 	imes \left(\frac{l}{2}ight) = \frac{3}{4} g$.

$A$ thin uniform stick of length $l$ and mass $m$ is held horizontally with its end $B$ hinged on the edge of a table. Point $A$ is suddenly released. The acceleration of the centre of mass of the stick at the time of release is:

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