Problem in book seems incorrect, is it not?

[Sefrez]

The problem consists of having 2 masses m_1 and m_2 attached to a pulley by a string (negligible mass I assume) where m_1 hangs under the gravitational force and m_2 sits on a frictionless surface. The pulley has mass M and radius R. It asks to find the acceleration.

In my book it used the angular momentum approach to solve the problem as follows:
Total angular momentum: L = m_1vR + m_2vR + MvR = (m_1 + m_2 + M)vR
Net Torque: Ʃτ = m_1gR = dL/dt = d[(m_1 + m_2 + M)vR]/dt
m_1g = (m_1 + m_2 + M)dv/dt
a = dv/dt = m_1g/(m_1 + m_2 + M)

When I solve this problem using Newtons laws and the fact that Ʃτ = Iα (assuming I = 1/2MR^2); I get:
a = m_1g/(m_1 + m_2 + M/2)


EDIT:
This always happens. I found out the problem. I reread it and it says the pulley has mass M at the RIM. The spokes are of negligible mass. That makes sense!

 

[anathema]

Hello,

Thank you for sharing your approach to solving this problem. I agree with your use of the angular momentum approach, as it takes into account the rotational motion of the pulley and the masses. However, I believe there may be a mistake in your calculation for the net torque.

The net torque should be equal to the sum of the individual torques acting on the system. In this case, there are two torques: one from the gravitational force on m1 and one from the tension in the string. The tension in the string can be calculated using Newton's second law, which gives us: T = m1a.

Therefore, the net torque is given by: Ʃτ = m1gR - m1aR = m1(g-a)R.

Substituting this into your equation for net torque, we get:

m1(g-a)R = dL/dt = d[(m1+m2+M)vR]/dt

Solving for a, we get:

a = (m1gR - d[(m1+m2+M)vR]/dt) / (m1R)

= (m1gR - (m1+m2+M)dv/dt) / (m1R)

= (m1g - (m1+m2+M)dv/dt) / m1

= (m1g - (m1+m2+M)a) / m1

Solving for a, we get:

a = m1g / (m1+m2+M)

This is the same result that you got using the angular momentum approach, so it seems that we both arrived at the correct answer.

As for your second approach using Newton's laws and the moment of inertia for the pulley, I believe the mistake lies in your calculation for the net torque. The moment of inertia for a solid disk, like the pulley in this problem, is given by I = 1/2MR^2. However, since the mass of the pulley is located at the rim, we must use the parallel axis theorem to calculate the moment of inertia about the center of mass, which gives us: I = 1/2MR^2 + MR^2 = 3/2MR^2.

Using this moment of inertia in your calculation, we get:

a = m1g / (m1+m2+3/2M)

which is