# Nonuniform Circular Motion-Find time given r, a-sub-t, V-sub-0

1. Sep 19, 2009

### 6maverick6

Nonuniform Circular Motion--Find time given r, a-sub-t, V-sub-0

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

A car turns through 60 degrees while traveling on the circumference of a 35m radius circle. The care is traveling at 8.0 m/s as it enters the turn and maintains a constant tangential acceleration of .4m/s2 throughout the turn. How long did it take the car to make the turn?

(A) 2.3s (B) 4.2s (C) 4.9s (D) 8.4s

DATA SUMMARY:
at: .4m/s2
Vo: 8.0m/s
R:35m

2. Relevant equations

at=d|v|/dt=.4m/s2

|V|= $$\omega$$ *r

$$\theta$$= $$\omega$$*r

3. The attempt at solution

at=d|v|/dt=.4m/s2

|V|=.4t +8.0 (from integrating at)

$$\omega$$= (.4t+8.0)/35 (from |V|=$$\omega$$*r)

$$\theta$$= $$\omega$$*r

($$\pi$$/3= ((.4t+8.0)/35)t= (.4/35) + (8.0t/35)

t=4.6

I get the feeling that there's one crucial equation I'm missing...

2. Sep 19, 2009

### Fightfish

Re: Nonuniform Circular Motion--Find time given r, a-sub-t, V-sub-0

$$\theta = \omega_{0} t + \frac{1}{2} \alpha t^{2}$$
should be the equation to be used instead.

$$\theta = \omega t$$ is true only under a constant angular velocity - the general form is $$\theta = \int \omega dt$$
which yields the equation above for constant angular acceleration.

You could also use a translational kinematics approach instead, which I think should be less confusing than rotational dynamics.

Last edited: Sep 19, 2009
3. Sep 19, 2009

### 6maverick6

Re: Nonuniform Circular Motion--Find time given r, a-sub-t, V-sub-0

Thank you!

I got it using translational kinematics.