Finding the tangential component of acceleration

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 6K views
jumbogala
Messages
414
Reaction score
4
EDIT: I meant radial in the title.

Homework Statement


A ball is going around in a circle of radius 4 m.

It goes with a constant angular velocity of (13 rad/s)[tex]\hat{k}[/tex] for 0.5 s. After that, it takes 4 s to come to a complete stop.

Find the radial component of the ball's acceleration at 2 s.

Homework Equations


The Attempt at a Solution


My book says that to use the formula ar= w2r. However, w is changing, so I don't see how I can use that!

The only thing I can think of is to find the angular acceleration:
[tex]\alpha[/tex] = w0 + [tex]\alpha[/tex]0(t)
0 = (13 rad/s) + [tex]\alpha[/tex]0(4 s). Solving for [tex]\alpha[/tex] gives -3.25 rad/s2[tex]\hat{k}[/tex]

Then I use another formula to find the angular velocity at 2 s:
wfinal = winitial + [tex]\alpha[/tex](t)
wf = (13 rad/s) + (-3.25 rad/s2)(2 s)
wf = 6.5 rad/s [tex]\hat{k}[/tex]

Then use that first formula:
ar = (6.5 rad/s)2(4 m)
ar = (169 rad/sm)[tex]\hat{k}[/tex]

Are those units correct? Really the formula for ar = dVt / dt, but is what I did ok?

Also, as an aside, the TANGENTIAL part of the angular acceleration would stay the same all the time, right? If I calculated it at 1 s, 2s, ... 4.3 s, it would not change?
 
Last edited:
Physics news on Phys.org
Hi jumbogala! :smile:

(have an alpha: α and an omega: ω :wink:)
jumbogala said:
… It goes with a constant angular velocity of (13 rad/s)[tex]\hat{k}[/tex] for 0.5 s. After that, it takes 4 s to come to a complete stop.

Find the radial component of the ball's acceleration at 2 s.

Your calculations are fine, except that you've misread the question …

you only have 1.5 s of acceleration at 2s. :wink:
Are those units correct? Really the formula for ar = dVt / dt, but is what I did ok?


You're right to be worried … the units in the formula v = ωr are cm/s = rad/s times cm … and in the formula a = ω2r are cm/s2 = rad2/s2 times cm … the radians are dimensionless, and they just drop out. :wink:
Also, as an aside, the TANGENTIAL part of the angular acceleration would stay the same all the time, right? If I calculated it at 1 s, 2s, ... 4.3 s, it would not change?

Not following this. :redface:

"tangential part of the angular acceleration" makes no sense.

Do you mean the tangential part of the ordinary acceleration (ie, the tangential acceleration)?

If so, then yes, you're correct … for fixed radius, that's simply dv/dt, the derivative of the speed (= r dω/dt = rα). :smile:
 
Thank you!

I'm confused about those units still, though. Why are we using cm, if the radius is given in m? Is that just a convention?

Also, the rad drops out for a, but if I just want to write ω, can I still write rad/s? (Instead of 1/s).
 
jumbogala said:
I'm confused about those units still, though. Why are we using cm, if the radius is given in m? Is that just a convention?

oh, I made a mistake … I thought the question used cm. :redface:
Also, the rad drops out for a, but if I just want to write ω, can I still write rad/s? (Instead of 1/s).

Yes, ω is rad/s. :smile: