# Homework Help: Angular Velo, Tangential Acceleration

1. Nov 28, 2008

### agross

I have computed angular speed, centripetal acceleration, and torque for a uniform circular motion.

What is the tangential acceleration?

I am leaning towards tangential a = r(alpha) where alpha is d(omega)/dt.
since I have calculated omega and a = dv/dt - I think its safe to
say that as dt cancels dv = r d(omega) = a tangential?

Any tips?

2. Nov 29, 2008

### horatio89

This is a very strange mathematical manipulation, because dt, is not actually an unknown by itself, and not directly "cancel-able". Tangential a, for uniform circular motion is given by r(omega)2 or v2/r depending on what information you have.

3. Nov 29, 2008

### alphysicist

This is not correct; for one thing the units are not the same (m/s^2 on the right hand side, m/s on the others).

$$\frac{dv}{dt} = r \frac{d\omega}{dt} = a_t$$

(where v is the speed along the circular path and $$a_t[/itex] is the tangential acceleration), then you can rewrite this as: [tex] dv = r\ d\omega = a_t\ dt$$

so the dt term is now associated with the tangential acceleration.

It's not clear from your post what other information you might have; but you can relate the torque to the total acceleration, and then relate total acceleration to the tangential acceleration.

I don't believe those equations are correct.

4. Nov 29, 2008

### horatio89

Ah, sorry, my mistake... I was using the equations for radial acceleration. Thanks for pointing it out.