# Centripetal acceleration car problem

1. Jan 4, 2007

### AznBoi

1. The problem statement, all variables and given/known data
A race car starts from rest on a circular track of radius 400m. The car's speed increases at the constant rate of .5m/s^2. At the point where the magnitudes of the centripetal and tangential accelerations are equal, determine a) the speed of the race car, b) the distance traveled, and c) the elapsed time.

2. Relevant equations
centripetial accel. =rw^2
tangential accel.= r(alpha)

3. The attempt at a solution
Hmm.. I'm a little bit confused at how to find the point where the two accelerations are the same. Well first I made the equations equal to each other and I tried solving for (t) but it seems like everything just cancels out except theta...

Another question, have you every been in a situation where you never understood anything that your class learned? How did you over come that? I mean I've listened to my teacher, read books and internet stuff, but I can't seem to get some facts straight no matter what. What did you do when you were totally stuck? I mean you didn't understand anything?

Thanks!

2. Jan 4, 2007

### Dorothy Weglend

Why make the accelerations equal to each other when the problem tells you what one of them is?

I think most of us are in that situation from time to time. I know I am. Just keep plugging away, ask questions, eventually it seems to sort itself out.

Best,
Dorothy

3. Jan 4, 2007

### tim_lou

you know that the linear speed of the car is constant, so from

$$a_{\text{tang}}=r\alpha$$

you will have a function of $\theta$ in terms of time.

4. Jan 5, 2007

### Hootenanny

Staff Emeritus
I don't think it is; (perhaps you meant the linear acceleration...)

Or am I missing something?

5. Jan 5, 2007

### Staff: Mentor

OK, but:
(1) The tangential acceleration is given, so use it.
(2) You can also write the centripetal acceleration in terms of tangential speed (instead of angular speed), which you can figure out from the tangential acceleration.

Show exactly what you did.

6. Jan 5, 2007

### AznBoi

Okay here is what I did lol:

Since a_c=rw^2 and a_t=r(alpha)

r(alpha)=rw^2

--Divide the radius from both sides

(alpha)=w^2

W/t=(theta/t)^2

Kept simplifying for some reason and ended up with t=(delta)theta

I wanted to solve for a variable,but how would you know what theta is? Or do you not use theta at all?

7. Jan 5, 2007

### AznBoi

Ok, so how do you know what is tangential acceleration when objects are moving in a circle? Do you look at the units (m/s^2) ? Centripetal acceleration is measured in rad/s^2 right?

If .5m/s^2 is the tangential acceleration, I think I can solve this. Let me try. =]

8. Jan 5, 2007

### AznBoi

Alright I've found both the angular speed and tangential velocity:

Since a_t=0.5m/s^2

0.5m/s^2=rw^2

0.5m/s^2=(400m)w^2

w=0.0354 rad/s^2 <---- Do you need to include (rad)? I've heard that it is a pure unit and can be counted as NOTHING. Is this true? If so, can I just write is as 0.0354 1/s^2 or should I include the (rad)?

Ok, a) says: determine the speed of the race car. Should I always assume that they are talking about the tangential speed and not the angular speed??
How do you know when they are talking about either one?

If they want the v, v=rw

v= 14.14m/s

Are my steps correctly done? Thanks for your help!

9. Jan 5, 2007

### AznBoi

Oops I think I made a mistake. =/

10. Jan 5, 2007

### AznBoi

Wait nvm. Since 0.5=a_t and a_t=r(alpha)

and the magnitudes of a_c and a_t are equal, then a_c=a_t right?

So, rw^2=.05m/s^2 correct?

11. Jan 5, 2007

### Staff: Mentor

The units of w are rad/s, not rad/s^2. Radians (and other angle measures) are an unusual unit, since angles have no real dimension: so give your answer for angular speed in terms of rad/s. But when you use w to find the tangential speed, you'll drop the radians.

You have to understand the context. If someone asked you "how fast" a car was going around a track, would you be tempted to give the angular speed? Of course not--they mean linear speed.
When they ask for speed, it generally means linear speed--as measured in m/s; but if they ask for angular speed, then that's what you give them.

Looks good to me.

You could save a bit of work by expressing the centripetal acceleration directly in terms of speed:
$$a_c = \omega^2 r = v^2/r$$

12. Jan 5, 2007

### AznBoi

Oh, ok yeah I don't know why I put ^2 with speed. lol. Ok, I get the (rad) units thing and when to use them now. I should remember that other equation, it could have been a lot faster like you said.

By the way, what is the relationship between v^2/r and rw^2? I know that a_c is the vector that is pointed towards the circular rotational motion and that it is can be caculated from v_f-v_i/t right? I don't see how you can change (omega) into tangetial velocity though.

I've done both b) and c) now using the speed I found. Here's my work:

Well I found time using the tangetial speed. + tangential accel

V=V_o+at

14.14m/s=0+(.5m/s^2)t

t=28.3s

b) asks for the distance traveled
I used x=1/2at^2

x=1/2(.5m/s^2)(28.3s)^2

x= 200.223m

c) asks for the elapsed time and I think I'v already found it!

It would be 28.3s right? Thanks for all your help doc al and everyone else!

13. Jan 5, 2007

### Hootenanny

Staff Emeritus
$$\omega = \frac{v^2}{r}$$

14. Jan 5, 2007

### Staff: Mentor

What do you mean you don't see how? You did it yourself in solving the first part!
$$v_t = \omega r$$

15. Jan 5, 2007

### AznBoi

I thought v=rw so how can w = v^2/r? Shouldn't it be w=v/r

EDIT: wait let me solve this since a_c=rw^2

16. Jan 5, 2007

### Staff: Mentor

Looks good to me!

17. Jan 5, 2007

### Staff: Mentor

Hoot made a typo. He is allowed one mistake per year--that's it for 2007.:rofl:

18. Jan 5, 2007

### Hootenanny

Staff Emeritus
My mistake, brain fart so to speak.

$$\alpha = \frac{v^2}{r}$$

$$\omega = \frac{v}{r}$$

As you correctly say.

The pressure is on now :rofl:

19. Jan 5, 2007

### AznBoi

Oh ok I got it now! =D a_c=rw^2 and v=rw so v/r=w, w^2=v^2/r^2

substitute that in a_c=rw^2 and you get a_c=r(v^2/r^2) =v^2/r

20. Jan 5, 2007

### AznBoi

lol. I'm sorry Hoot. Okay, thanks to both of you especially and everyone else that helped me!