# From torque to acceleration

1. Nov 19, 2008

### saltine

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

Find acceleration a given the torque, T, about the axle.

2. Relevant equations
$$\tau = I\frac{d\omega}{dt} = I\alpha$$

Moment of Inertia of a disk (about the axle):
$$I_z = \frac{1}{2}MR^2$$

3. The attempt at a solution

$$\alpha = Ra$$ in this context, so
$$\tau = I Ra$$
$$a = \frac{\tau}{IR} = \frac{2\tau}{MR^3}$$
I think this is correct, however:

$$\tau = R \times F$$, since F = ma, $$a = \frac{\tau}{mR}$$.
What does this a represent? Does it represent the linear acceleration of a particle of mass m that happens to be on the edge of the disk?

- Thanks

2. Nov 20, 2008

### tiny-tim

Welcome to PF!

Hi saltine! Welcome to PF!

(i assume the wheel rolls without slipping?)
No, $R\alpha = a$ … just consider the dimensions!
No, I don't understand that … tau doesn't come from a force at a distance R … it probably comes from two forces very close together, on either side of the axle.