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Ahat [919]
3 years ago
10

An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2.3 kg mass of the pulley is concentrated on i

ts 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?
Physics
1 answer:
Sergeu [11.5K]3 years ago
5 0

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}

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