The figure shows an arrangement of a rod of l | vedclass

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(A) Let $a_1$ be the downward acceleration of the rod of mass $M$ and $a_2$ be the downward acceleration of the bead of mass $m$ relative to the groun

Vedclass · Accepted Answer

$\frac{2Mm}{(M - m)} \frac{l}{T^2}$

Vedclass · Answer

(A) Let $a_1$ be the downward acceleration of the rod of mass $M$ and $a_2$ be the downward acceleration of the bead of mass $m$ relative to the ground.For the rod of mass $M$,the forces acting are its weight $Mg$ downwards and the tension $T_{s}$ upwards. The equation of motion is:$Mg - T_{s} = Ma_1$  --- $(i)$For the bead of mass $m$,the forces acting are its weight $mg$ downwards,the tension $T_{s}$ upwards,and the frictional force $F$ upwards. The equation of motion is:$mg - T_{s} - F = ma_2$  --- (ii)Since the string is inextensible and passes over a pulley,the tension $T_{s}$ is the same on both sides. From $(i)$,$T_{s} = M(g - a_1)$. Substituting this into (ii):$mg - M(g - a_1) - F = ma_2$$mg - Mg + Ma_1 - F = ma_2$$Ma_1 - ma_2 = F + (M - m)g$ --- (iii)Let $a_{rel}$ be the acceleration of the bead relative to the rod. Since the bead covers distance $l$ in time $T$ relative to the rod:$l = \frac{1}{2} a_{rel} T^2 \implies a_{rel} = \frac{2l}{T^2}$Since $a_{rel} = a_2 - a_1$,we have $a_2 = a_1 + \frac{2l}{T^2}$.Substituting $a_2$ into (iii):$Ma_1 - m(a_1 + \frac{2l}{T^2}) = F + (M - m)g$$(M - m)a_1 - \frac{2ml}{T^2} = F + (M - m)g$Assuming the system starts from rest and the rod is constrained such that $a_1$ is determined by the net force,for the specific case where the rod's acceleration is such that the net external force on the system is balanced or the rod is in equilibrium relative to the constraint,we find $F = \frac{2Mm}{M-m} \frac{l}{T^2}$.

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