A mass m₁ is connected by a light string that passes over a pulley of mass M to a mass m₂ sliding on a frictionless horizontal surface as shown in the figure.There is no slippage between the string and the pulley.The pulley has a radius of 25.0 cm and a moment of inertia of 1/2 MR².If m₁ is 4.00 kg,m₂ is 4.00 kg,and M is 4.00 kg,then what is the acceleration of m₁? A) 4.42 m/s² B) 3.92 m/s² C) 3.42 m/s² D) 3.04 m/s² E) 2.96 m/s²

Question

Quizplus · Accepted Answer

We can start by applying Newton's second law to each mass.
For m1: m1g-T= m1a
For m2: T= m2a

Since the string is not slipping over the pulley, we know that the acceleration of the pulley is equal to the acceleration of m1 and m2. We can use this to get the torque equation for the pulley.

Torque= I*alpha
T*(2R)= (1/2)M*(R^2)*alpha
a= alpha*R

Substituting the expression for T in terms of a, we get:
m2a*(2R)= (1/2)M*(R^2)*(a/R)
Simplifying, we get:
a= (2/3)*(m2/m1+M)*g

Plugging in the given values, we get:
a= (2/3)*(4/(4+4+4))*9.8= 3.92 m/s^2

A Mass M1 Is Connected by a Light String That

A Mass M₁ Is Connected by a Light String That