Angular Velo, Tangential Acceleration

[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?

 

[horatio89]

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?

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.

 

[alphysicist]

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?

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).

If you start with:

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

(where v is the speed along the circular path and [tex]a_t[/itex] is the tangential acceleration), then you can rewrite this as:<br /> <br /> $$dv = r\ d\omega = a_t\ dt$$<br /> <br /> so the dt term is now associated with the tangential acceleration.<br /> <br /> <br /> 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.<br /> <br /> <br /> <br /> <br /> <br /> <blockquote data-attributes="" data-quote="horatio89" data-source="post: 1979416" cite="https://www.physicsforums.com/goto/post?id=1979416" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> horatio89 said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 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. </div> </div> </blockquote><br /> <br /> I don't believe those equations are correct.[/tex]

 

[horatio89]

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