Question

A bucket of mass m is tied to a massless cable that is wrapped around the outer rim of a frictionless uniform pulley of radius R, similar to the system shown in Fig. E9.47. 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?

Answer 1

To determine the moment of inertia of the pulley so that it always has half as much kinetic energy as the bucket, you can equate the kinetic energy of the pulley to half the kinetic energy of the bucket. Let the mass of the bucket be m and the moment of inertia of the pulley be I. The equation to solve for I is I = (m * v^2) / ω^2.

Step-by-step explanation:

To determine the moment of inertia of the pulley so that it always has half as much kinetic energy as the bucket, we need to equate the kinetic energy of the pulley to half the kinetic energy of the bucket.

Let the mass of the bucket be m. The kinetic energy of the bucket is given by:

Kinetic energy of bucket = 1/2 * m * v^2

where v is the velocity of the bucket.

Let the moment of inertia of the pulley be I and the angular velocity of the pulley be ω. The kinetic energy of the pulley is given by:

Kinetic energy of pulley = 1/2 * I * ω^2

Since the pulley always has half as much kinetic energy as the bucket, we can equate the two expressions and solve for I:

1/2 * I * ω^2 = 1/2 * m * v^2

I = (m * v^2) / ω^2

Answer 2

The moment of inertia of the pulley must be (m * v²) / (2 * omega²) to always have half as much kinetic energy as the bucket.

Step-by-step explanation:

The moment of inertia of the pulley can be determined by setting up an equation using the work-energy theorem and the given information. The work-energy theorem states that the work done on an object is equal to its change in kinetic energy. In this case, the work done on the pulley is equal to the change in its rotational kinetic energy. Half of the kinetic energy of the bucket is equal to the change in kinetic energy of the pulley. Therefore, we can set up the equation:

1/2 * m * v² = delta(1/2 * I * omega²)

Where m is the mass of the bucket, v is the velocity of the bucket, I is the moment of inertia of the pulley, and omega is the angular velocity of the pulley.

Solving for I, we get:
I = (m * v²) / (2 * omega²)

In terms of the given variables, the moment of inertia of the pulley must be (m * v²) / (2 * omega²) to always have half as much kinetic energy as the bucket.