# Rotational motion question

1. Mar 8, 2010

### sheepcountme

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

Two cars race around a circular track. Car A accelerates at 0.340 rad/s2 around the track, and car B at 0.270 rad/s2. They start at the same place on the track and car A lets the slower-to-accelerate car B start first. Car B starts at time t = 0. When car A starts, car B has an angular velocity of 1.40 rad/s. At what time does car A catch up to car B?

2. Relevant equations
rotational motion equations

3. The attempt at a solution

I attempted to use rotational motion equations and set up

theta car A=theta car B
since they will be in the same place (theta) at the same time.

as in: omega initial A x time + 1/2 alpha A time^2 = omega initial B x time + 1/2 alpha A time^2

So: 0(t) + 1/2(.340)t^2=1.4(t)+1/2(.270)t^2
and then I solved for time using quadratic theorem and got 40 seconds but this was incorrect. Could you tell me where I went wrong?? Thanks!

2. Mar 8, 2010

### aim1732

Is the angle really equal ? The second car already has traversed some angle when the other starts. So angle by faster car > angle by slower car.

3. Mar 8, 2010

### dpeagler

You first have to find delta theta to see how far car B has gone. Then find how long it took to traverse that distance. Then plug your delta theta into the equations for car A with its respective angular velocity then add the two separate times (first time while car A was waiting + time it took car A to catch up) should be way less than 40 seconds if I'm not mistaking.

4. Mar 8, 2010

### sheepcountme

Okay, so I used omegafinal^2=omegainitial^2+2(alpha*deltatheta) to find delta theta for car B coming to delta theta=3.63rad

Then omegafinal=omegainitial+alphat to find the time for car B to reach this point, getting t=5.19s

Then I set the previous delta theta equal to an equation for Car A:
(3.63)=omegainitialt+1/2(alpha)t^2
3.63=0+1/2(.34)t^2
and solved for t getting 4.62 seconds

and so I added this 4.62 to 5.19 getting 9.81 however this was incorrect...

5. Mar 8, 2010

### dpeagler

The first two parts you did are correct ( when you found the radians and the time ), as far as I can tell.

You didn't put your Theta initial B as 3.63 radians. This should fix your problem. Sorry I haven't had time to sit down and work it out to make sure this is the mistake, but I'm pretty sure.

6. Mar 10, 2010

### sheepcountme

So, using the equation above, I get
1/2(.340)t^2=3.63t+1/2(.270)t^2
0=-.035t^2+3.63t

and time ends up being 103.72s This seems much too big...I hate needing someone to hold my hand through this, but there's really something I'm missing.

7. Mar 10, 2010

### dpeagler

You're still not using that equation correctly.

$$\theta$$= $$\theta$$$$_{int}$$ + $$\omega$$$$_{int}$$t + $$\frac{1}{2}$$$$\alpha$$t$$^{2}$$

(sorry about the formatting still trying to get used to LaTeX.)

You then set them equal to each other. You put your radians in the wrong place on the equation you listed. You put them where the initial angular velocity goes. This should bring your answer down by around a factor of two.