# Acceleration of falling mass and tension in the cord

1. Dec 4, 2011

### skysunsand

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

A 25 kg wheel has a radius of 40cm and turns freely on a horizontal axis. The radius of gyration is 30cm. A 1.2 kg mass hangs at the end of a cord that is wound around the rim of the wheel. This mass falls and causes the wheel to rotate. Find the acceleration of the falling mass and the tension in the cord

2. Relevant equations

I= Mk^2
a= $\alpha$ r

3. The attempt at a solution

I figured out the inertia, with that being I=25*.3^2 to get 2.26.

The downward force on the object would be mg-Ft
The upward force is just Ft, I believe.

But somewhere in there has to go acceleration, which is a=$\alpha$*.4

I think I need another equation to set everything to, but I don't know what to do...

2. Dec 4, 2011

### grzz

eqn 1.............F(net) = ma for the falling mass

eqn 2 ...........torque = Iα for the rotating disk

a = Rα

3. Dec 4, 2011

### skysunsand

Why do I need torque?

4. Dec 4, 2011

### grzz

Without the torque equation you will end up with too many unknown quantities.

5. Dec 4, 2011

### skysunsand

But torque is another unknown.

I have

Ft= 25*a

and

Torque = 2.25 * angular acceleration

And I have a=.4*angular acceleration

So if I solve for either of them, I still wind up with two unknowns in the equation.

6. Dec 4, 2011

### grzz

What is this equation?

7. Dec 4, 2011

### skysunsand

Force of the tension = mass times acceleration.

Should it be mass of the object?

so Ft= 1.2a

But either way, that doesn't solve my too many variables problem

8. Dec 4, 2011

### grzz

F(net) = ma...............for falling mass

But F(net) downwards = mg - Tension

therefore

mg - Tension = ma

1.2 x 9.8 - tension = 1.2a

In above we have still 2 unknows

But we also have torque = I x ang acc and linear acc a - R x ang acc