# Pulley Problem Dealing with Torque

1. Apr 5, 2012

### PeachBanana

1. The problem statement, all variables and given/known data

A 1.53-kg mass hangs on a rope wrapped around a frictionless disk pulley of mass 7.07 kg and radius 66.0 cm. What is the acceleration of the mass?

I know this has been posted before but I learn better when I try to solve things on my own.

2. Relevant equations

F = ma
T = Iα

3. The attempt at a solution

Forces acting on mass: unknown acceleration and torque from pulley
Forces acting on disk : Gravity

So there are two forces pointing down (gravity on the mass and disk) and one pointing up (torque).

(1.53 kg)(unknown acceleration) + (7.07 kg) (9.8 m/s^2) = 1/2 (0.66 m)^2 (7.07 kg)

Does this make sense? I want to ensure I've labeled forces correctly.

2. Apr 5, 2012

### Staff: Mentor

Not quite. For one thing, acceleration and torque are not forces.

Treat the hanging mass and the disk separately. Start with the hanging mass: What forces act on it?

3. Apr 5, 2012

### PeachBanana

Gravity acts on the hanging mass. (1.53 kg)(9.8 m/s^2) ≈ 15 N. There's also an upward force that acts on the hanging mass that prevents it from free falling. (1.53 kg)(a m/s^2).

Gravity also acts on the disk too. (7.07 kg)(9.8 m/s^2) = 69.3 N. Now I'm unsure of what to do with the 0.66 m.

4. Apr 5, 2012

### Staff: Mentor

Two forces act on the hanging mass: Gravity (mg) acting down and the tension in the rope acting up. Write an equation for Newton's second law for the hanging mass. You'll have two unknowns, the tension and the acceleration.
It's true that gravity acts on the disk, but we don't care because it doesn't exert a torque on the disk. But the rope tension does. Write an equation for the rotational form of Newton's 2nd law for the disk.

5. Apr 5, 2012

### PeachBanana

Ok. Tension force = (7.07 kg)(some acceleration)
This force points up.

Torque = 1/2 (7.07 kg)(0.66 m)^2 (α)
This is produced on the hanging mass.

F = (1.53 kg)(a)

I have three equations with three unknowns. Should I be rewriting "α" differently?

6. Apr 5, 2012

### Staff: Mentor

What you wrote is "ma". That's not the tension, but the effect of the net force via Newton's 2nd law. Don't try to guess the tension, just call it T. You'll solve for it if you need it.

Use ƩF = ma. What's ƩF?

Again, what you wrote is "Iα". That's the effect of the net torque as given by Newton's 2nd law for rotation.

Use ƩTorque = Iα. What's the torque on the disk? (Write it in terms of the forces acting on the disk.)

The linear acceleration (a) and the angular acceleration (α) are related. How?

7. Apr 5, 2012

### PeachBanana

I think I'm getting closer.

1. $\Sigma$F = T - ma

2. Torque on disk: F * r

3. r * α = a

F = (7.07 kg)(9.8 m/s^2) - (1.53 kg)(a)

Torque on disk = (7.07 kg)(9.8 m/s^2)(0.66 m)

(0.66 m)(a) = a

8. Apr 5, 2012

### Staff: Mentor

First find the net force on the hanging mass. The only forces are mg and T, so what's the net force? (Take down to be positive, since that's the direction of the acceleration.)

Then set that net force equal to "ma".
What's F? The torque on disk is due to the tension, so: T*r.

Set that torque equal to I*α.
Good!