How Does a Frictionless Pulley Influence Acceleration in Connected Masses?

[123yt]

Pretend there are two accelerating masses connected to a massless string with a frictionless pulley between them. How can the frictionless pulley (Rotational inertia and radius given) affect acceleration in any sort of way?

Also, why is the net force equal to Acceleration * (Mass of two blocks + I/r^2)? I understand the part with the two blocks, but not with the I/r^2.

 

[Doc Al]

Pretend there are two accelerating masses connected to a massless string with a frictionless pulley between them. How can the frictionless pulley (Rotational inertia and radius given) affect acceleration in any sort of way?

The pulley has rotational inertia and thus requires a torque to accelerate it.

Also, why is the net force equal to Acceleration * (Mass of two blocks + I/r^2)? I understand the part with the two blocks, but not with the I/r^2.

You can think of I/r^2 as the effective mass of the pulley. But that equation is a bit of a short cut. Rather than use it directly, derive your own version by applying Newton's 2nd law to each mass and the pulley itself.

 

[123yt]

The pulley has rotational inertia and thus requires a torque to accelerate it.

But torque is just a measure of how much a force causes an object to rotate. It doesn't "use up" any force to rotate it, right?

You can think of I/r^2 as the effective mass of the pulley. But that equation is a bit of a short cut. Rather than use it directly, derive your own version by applying Newton's 2nd law to each mass and the pulley itself.

The rotational inertia of the pulley is I = MR^2 / 2, so shouldn't the mass be M = 2 * I / R^2?

 

[Doc Al]

But torque is just a measure of how much a force causes an object to rotate. It doesn't "use up" any force to rotate it, right?

It "uses up" force in a manner similar to how pushing a mass "uses up" force.

The rotational inertia of the pulley is I = MR^2 / 2, so shouldn't the mass be M = 2 * I / R^2?

No. If you derive the equation, you'll see where that I/R^2 term comes from. (No reason to treat the pulley as a uniform disk.)

 

[123yt]

Alright, thanks for the help. I think I understand torque and rotation a little better now.