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The discussion centers on a physics problem involving a light, flexible cable wrapped around a hollow cylinder, which rotates as a block of mass m falls. The key point is that the speed of the block just before it hits the floor is derived using energy conservation principles. The formula for the speed is v = sqrt(2mgh/(m+M)), which differs from the speed of a solid cylinder, given by v = sqrt(2gh/(1+M/2m)). The difference arises from the distribution of mass in the two cylinder types, affecting the moment of inertia and thus the rotational kinetic energy. In the case of the hollow cylinder, more energy is allocated to rotation compared to the solid cylinder, leading to a lower speed of the falling mass. This highlights the importance of considering the moment of inertia in rotational dynamics when analyzing systems involving both linear and rotational motion.
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We wrap a light, flexible cable around a thin-walled, hollow cylinder with mass M and radius R. The cylinder is attached to the axle by spokes of a negligible moment of inertia.The cylinder rotates with negligible friction about a stationary horizontal axis. We tie the free end of the cable to a block of mass m and release the object with no initial velocity at a distance h above the floor. As the block falls, the cable unwinds without stretching or slipping, turning the cylinder. The speed of hanging mass m just as it strikes the floor is v=sqrt 2mgh/(m+M). Use energy concepts to explain why the answer to part A is different from the speed found in case of solid cylinder,which is
v= sqrt 2gh/(1+M/2m) ?
 
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