A mass (M₁ = 5.0 kg)is connected by a light cord to a mass (M₂ = 4.0 kg)which slides on a smooth surface,as shown in the figure.The pulley (radius = 0.20 m)rotates about a frictionless axle.The acceleration of M₂ is 3.5 m/s².What is the moment of inertia of the pulley? A)0.29 kg.m² B)0.42 kg.m² C)0.20 kg.m² D)0.62 kg.m² E)0.60 kg.m²

Question

Quizplus · Accepted Answer

We can use the formula for rotational motion, τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. We know that the torque is given by τ = F*r, where F is the force and r is the radius of the pulley. We can assume that the tension in the cord is the same on both sides and use Newton's second law to find the force on the system:

F = (m1 - m2)*a
F = (5.0 kg - 4.0 kg)*(3.5 m/s^2)
F = 3.5 N

Now, we can find the torque:

τ = F*r
τ = 3.5 N*0.20 m
τ = 0.7 N*m

Finally, we can find the moment of inertia:

τ = Iα
I = τ/α
I = τ/(a/r)
I = 0.7 N*m/(3.5 m/s^2)/(0.20 m)
I = 0.20 kg*m^2

Therefore, the moment of inertia of the pulley is 0.20 kg*m^2, which corresponds to choice C.

A Mass (M1 = 5

A Mass (M₁ = 5