# A ball of mass m is tied to a string that is wrapped around the outside of a pulley. The pulley, which is a uniform solid disk, also has a mass m, and the pulley rotates without friction about the axle through the middle of thepulley. After the system is released from rest, what is the acceleration of the ball?

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This image has been Flagged as inappropriate Click to unflag Let's start by writing down Newton's second law for the ball:

`mvec(g) + vecT = mveca` : The gravity and the force from the string on the ball, `vecT` , provide the ball's acceleration. This force is directed upward, and the gravity is directed downward. If we assume that the...

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Let's start by writing down Newton's second law for the ball:

`mvec(g) + vecT = mveca` : The gravity and the force from the string on the ball, `vecT` , provide the ball's acceleration. This force is directed upward, and the gravity is directed downward. If we assume that the acceleration is also downward, the equation becomes

`mg - T = ma`

The force from the string on the ball is equal in magnitude to the tension in the string, and it is also equal in magnitude to the force from the string on the pulley. This force provides the torque which is responsible for the rotation of the pulley. The magnitude of the torque, calculated with respect to the pulley's center, is

`tau = T*r` (Since `vecT` is tangential, it is perpendicular to the pulley's radius.)

Newton's second law for rotation is

`I*beta = tau` , where `beta` is angular acceleration and I is the moment of inertia of the pulley. The moment of inertia of the pulley, which is a solid disk, around its central axis is

`I = mr^2/2`

Plugging this into the equation above, we get

`mr^2/2 * beta = Tr`

The angular acceleration is related to the linear acceleration of a point on the rim of the pulley as `beta = a/r` . The linear acceleration of a point on the rim of the pulley is the same as the acceleration of the ball, because they are connected by the string. (If the acceleration were different, the string would break.)

So the equation becomes

`mr^2/2*a/r = Tr`

From here

`T = ma/2`

Combining this with Newton's second law, which we wrote down in the beginning, we get

`mg - ma/2 = ma`

From here `mg = 3ma/2`  and `a = 2/3g`

The answer, then, is: a = 2g/3.

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