A bucket of mass m is tied to a massless cable that is wrapped around the outer rim of a uniform pulley of radius R , on a frictionless axle, as shown in (Figure 1). In terms of the stated variables, what must be the moment of inertia of the pulley so that it always has half as much kinetic energy as the bucket? Express your answer in terms of the variables m and R .

Respuesta :

Answer:

I = ½ mR²

Explanation:

Translational kinetic energy of the bucket is KE = ½ mv².

Rotational kinetic energy of the pulley is RE = ½ Iω².

Since the pully has half the energy of the bucket:

RE = ½ KE

½ Iω² = ½ (½ mv²)

Iω² = ½ mv²

Since there's no slipping, v = ωR.

Iω² = ½ m (ωR)²

I = ½ mR²

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