A rod of mass ‘M’ and length $$6R$$ is pivoted at its centre (point$$ $$P) to rotate in vertical plane as shown. Two discs of same radius ‘R’ are joined at the ends $$O1$$ and $$O2$$ of the rod. Mass of the discs A and B are ‘M’ and ‘$$2M$$’ respectively. Initial angular velocities of discs are shown in figure. The system is released from rest from horizontal position. The angular velocity of the rod when it is vertical is gλR . Find the value of λ.

Question

A rod of mass ‘M’ and length $$6R$$ is pivoted at its centre (point$$ $$P) to rotate in vertical plane as shown. Two discs of same radius ‘R’ are joined at the ends  $$O1$$ and  $$O2$$ of the rod. Mass of the discs A and B are ‘M’ and ‘$$2M$$’ respectively. Initial angular velocities of discs are shown in figure. The system is released from rest from horizontal position. The angular velocity of the rod when it is vertical is gλR . Find the value of  λ.

Infinity Learn · Accepted Answer

Since, all forces acting on discs are passing through their centers, angular speeds of discs will not change.From conservation of energychange in potential energy $$=$$ Rotational kinetic energy of the rod ⇒∆$$PE$$$$=$$ Itotal ω$$22$$      Where Itotal $$=$$ total moment of inertia of the system⇒(2Mg3R)−$$(Mg3R)=12M(6R)212+M(3R)2+2M{3R}2$$ω$$2$$⇒$$3MgR=1230MR2$$ω$$2=15MR2$$ω$$2$$⇒$$3MgR=15MR2$$ω$$2$$⇒ω$$=g5R$$

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