A uniform rod of mass m and length ell hinged | vedclass

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(B) The torque about the hinge $A$ is given by $	au_A = I_A \alpha$. Since the rod is uniform,the weight $mg$ acts at the center of mass $C$ (at dist

Vedclass · Accepted Answer

$A \rightarrow Q, B \rightarrow Q, C \rightarrow R, D \rightarrow P$

Vedclass · Answer

(B) The torque about the hinge $A$ is given by $	au_A = I_A \alpha$. Since the rod is uniform,the weight $mg$ acts at the center of mass $C$ (at distance $\ell/2$ from $A$). $	au_A = mg 	imes \frac{\ell}{2}$. The moment of inertia of the rod about the hinge $A$ is $I_A = \frac{m\ell^2}{3}$. Equating the two: $mg 	imes \frac{\ell}{2} = \frac{m\ell^2}{3} \alpha$. Solving for angular acceleration: $\alpha = \frac{3g}{2\ell}$. This angular acceleration $\alpha$ is the same for all points on the rod. Thus,$(A) ightarrow Q$ and $(B) ightarrow Q$. The linear acceleration of any point at distance $r$ from the hinge is $a = \alpha r$. For point $C$ $(r = \ell/2)$: $a_C = \alpha 	imes \frac{\ell}{2} = \frac{3g}{2\ell} 	imes \frac{\ell}{2} = \frac{3g}{4}$. Thus,$(C) ightarrow R$. For point $B$ $(r = \ell)$: $a_B = \alpha 	imes \ell = \frac{3g}{2\ell} 	imes \ell = \frac{3g}{2}$. Thus,$(D) ightarrow P$. Therefore,the correct matching is $A ightarrow Q, B ightarrow Q, C ightarrow R, D ightarrow P$.

Column $I$	Column $II$
$(A)$ Angular acceleration of $C$	$(P)$ $\frac{3g}{2}$
$(B)$ Angular acceleration of $B$	$(Q)$ $\frac{3g}{2\ell}$
$(C)$ Acceleration of $C$	$(R)$ $\frac{3g}{4}$
$(D)$ Acceleration of $B$	$(S)$ $\frac{3g}{\ell}$

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