## 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

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?

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1 year 2021-09-04T04:43:07+00:00 1 Answers 40 views 0

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

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

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