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A block of mass m is kept on a horizontal table at a distance r from the centre of table the coefficient of friction between the turntable and block is my now the turn tablestarts rotating with uniform accelerationalpha the net acc.as a function time?

This question is a little vague as to what it is asking, but here's how the scenario plays out. There is a maximal frictional force of μ(Fn), where Fn is the normal force, in this case the weight, mg. As the turntable begins to slowly spin, the centrifugal force needed to keep it at a radius r increases, and this is provided by the friction. I assume this is the portion of the motion you refer to, as after this maximal friction is exceeded you need to consider radial changes too. During this portion of the motion, while held at a radius r, the angular velocity at time t is given by at, where a is the alpha given in the question. Therefore the central force required to fix that radius, as provided by the friction, is mrΩ^2, where Ω=ang. Velocity= at Therefore the central force is mr(a^2)(t^2). By Newton's second law, force=mass X acceleration, we conclude that the acceleration is r(a^2)(t^2)

Damon Marlow
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643 students helped

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