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A smooth rod of length l rotates freely in a horizontal plane with the angular velocity about a stationary vertical axis O as shown in Fig. The moment of inertia of the rod is equal to I relative to the axis. A small ring of mass m is located on the rod close to the rotation axis and is tied to it by a thread. When the thread is burned, the ring starts sliding radially outwards along the rod. Find the velocity v of the ring relative to the rod as a function of its distance r from the rotation axis. |
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Answer» SOLUTION :Consider the rod and the RING as our system. The system is experiencing no external torque about the axis hence, the angular momentum relative to the rotation axis does not VARY. There is no nonconservative force or IMPULSE acting so in the process of motion, the kinetic energy of the system does not vary. Calculation: Since the angular momentum about the axis and kinetic energy of the system are conserved, we have `Iomega_(0)=(I+mr^(2))omega` and `1/2Iomega_(0)^(2)=1/2Iomega^(2)+1/2mv^(2)` where `v^(2)=v.^(2)+omega^(2)r^(2)` as shown in Fig. From the above two equations we obtain `v=omega_(0)r//sqrt(1+mr^(2)//I)` Note: If r = 0, we get v = 0 which is the situation in the beginning. 
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