- #1

vparam

- 17

- 3

- Homework Statement
- Find the linear acceleration of an Atwoods machine with a pulley with moment of inertia I and radius R with masses m1 and m2 attached on either end.

- Relevant Equations
- ∑τ = Iα

We solved this problem in class as follows:

Net torque about the center of the pulley taking counterclockwise rotation to be positive = m1gR - m2gR = I_tot α, where I_tot is the moment of inertia of the full system.

My professor said that I_tot = I + m1R^2 + m2R^2, where m1R^2 is the moment of inertia of mass 1 about the center of the pulley (assuming it is a point mass). However, I don't see why we can use R here rather than the distance from the center of the pulley to the mass. Isn't the moment of inertia = mr^2, where r is the distance the center of the pulley to the position of m1? This r ≥ R by the Pythagorean theorem, so I'm not sure where this comes from since the moment of inertia doesn't have a cross product in the formula. In a disk for example, when integrating I = ∫r^2 dm, we use the radial distance from the center, not this R which seems to be the horizontal component of r.

I resolved the problem using angular momentum (and verified with the textbook), and got the same answer as the professor, so it seems that his method does work though. So, where am I going wrong with my understanding of moment of inertia and this value of r?

Net torque about the center of the pulley taking counterclockwise rotation to be positive = m1gR - m2gR = I_tot α, where I_tot is the moment of inertia of the full system.

My professor said that I_tot = I + m1R^2 + m2R^2, where m1R^2 is the moment of inertia of mass 1 about the center of the pulley (assuming it is a point mass). However, I don't see why we can use R here rather than the distance from the center of the pulley to the mass. Isn't the moment of inertia = mr^2, where r is the distance the center of the pulley to the position of m1? This r ≥ R by the Pythagorean theorem, so I'm not sure where this comes from since the moment of inertia doesn't have a cross product in the formula. In a disk for example, when integrating I = ∫r^2 dm, we use the radial distance from the center, not this R which seems to be the horizontal component of r.

I resolved the problem using angular momentum (and verified with the textbook), and got the same answer as the professor, so it seems that his method does work though. So, where am I going wrong with my understanding of moment of inertia and this value of r?