Question

An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2.3 kg mass of the pulley is concentrated on its rim which is a distance 23.5 cm from the axle. The mass on the right is 1 kg and on the left is 1.65 kg.
What is the magnitude of the linear acceleration a of the hanging masses?

Answer 1

Answer:

$a = 2.77~{\rm m/s^2}$

Explanation:

Since the pulley has a mass concentrated on its rim, the pulley can be considered as a ring.

The moment of inertia of a ring is

$I = mr^2 = (2.3)(23.5\times 10^{-2})^2 = 0.127$

The mass on the left is heavier, that is the pulley is rotating counterclockwise.

By Newton's Second Law, the net torque is equal to moment of inertia times angular acceleration.

$\tau = I \alpha$

Here, the net torque is the sum of the weight on the left and the weight on the right.

$\tau = m_1gR - m_2gR = (1.65)(9.8)(23.5\times 10^{-2}) - (1)(9.8)(23.5\times 10^{-2}) = 1.497~{\rm Nm}$

Applying Newton's Second Law gives the angular acceleration

$\tau = I\alpha\\1.497 = 0.127\alpha\\\alpha = 11.78~{\rm rad/s^2}$

The relation between angular acceleration and linear acceleration is

$a = \alpha R$

Then, the linear acceleration of the masses is

$a = 11.78 \times 23.5\times 10^{-2} = 2.77~{\rm m/s^2}$