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(C) Let $M$ be the mass of the metre stick. The centre of mass of the metre stick is at the $50.0 \,cm$ mark.When the stick is balanced at the $46.0 \

Vedclass · Accepted Answer

$72$

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(C) Let $M$ be the mass of the metre stick. The centre of mass of the metre stick is at the $50.0 \,cm$ mark.When the stick is balanced at the $46.0 \,cm$ mark, the torque due to the weight of the stick must be balanced by the torque due to the coins.The weight of the four coins is $W_c = 4 	imes 2 \,g = 8 \,g$. This acts at the $10.0 \,cm$ mark.The distance of the coins from the pivot $(46.0 \,cm)$ is $d_1 = 46.0 \,cm - 10.0 \,cm = 36.0 \,cm$.The weight of the metre stick $W_s = M \,g$ acts at its centre of mass, which is at the $50.0 \,cm$ mark.The distance of the centre of mass from the pivot is $d_2 = 50.0 \,cm - 46.0 \,cm = 4.0 \,cm$.For rotational equilibrium, the clockwise torque must equal the counter-clockwise torque:$M \,g 	imes d_2 = W_c 	imes d_1$$M \,g 	imes 4.0 \,cm = 8 \,g 	imes 36.0 \,cm$$4 \,M = 8 	imes 36$$M = 2 	imes 36 = 72 \,g$.

$A$ metre stick is balanced on a knife edge at its centre. When four coins, each of mass $2 \,g$, are placed one on top of the other at the $10.0 \,cm$ mark, the stick is found to be balanced at the $46.0 \,cm$ mark. The mass of the metre stick is: (in $\,g$)

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