# Angular Acceleration

1. Dec 6, 2009

### LBenson

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

A motor car with wheels of 0.6,m diameter accelerates uniformly from 3 to 18m/s in 10 seconds.

Find:

a) the angular acceleration of the wheels
b) the number of revolutions made by each wheel during the speed change

2. Relevant equations

A=v-u/t (angular equivilent with Alpha as acceleration etcetc)

3. The attempt at a solution

My first thought was to change the 3m/s and 18m/s into Rad/s but i'm unsure how to go about doing that.

Then using A=v-u/t i assume the acceleration can be found

I have no clue how to attempt the second part.

Cheers.

2. Dec 6, 2009

### Oddbio

You can relate the linear speed and acceleration with the rotational values by realizing that the tires should be in contact with the ground the whole time, and so any distance covered by the vehicle will be equal to the arc length that the wheels have rotated by.
Similarly any acceleration in the vehicle will correspond to an "acceleration" (or change) of the arc length.
If s=arc length r=radius and θ=the angle rotated
then use the relation:
s=rθ

3. Dec 6, 2009

### LBenson

Sorry, i'm a tad lost over this whole situation...

I really have no clue where to start

4. Dec 6, 2009

### ApexOfDE

I think, you should consider an equation that show relationship between velocity and angular velocity. Then derive it and from here you can find angular acceleration.

5. Dec 6, 2009

### LBenson

Ok i think i got it.

After looking back through my notes i found V=wr (W being angular velocity)

I rearranged for w (w=v/r) and worked out the angular velocity for both the 3m/s and the 18m/s

So w1=3/0.3 (since 0.6 was the diameter 0.3 = radius) = 10rad/s

Then
A=w2-w1/t

Which is the right answer since i have the marks scheme infront of me.

Now just to crack on with the next part. Cheers for the help.

6. Dec 6, 2009

### LBenson

Should have refreshed my page a few minutes earlier :)

Thanks for the help.

7. Dec 6, 2009

### LBenson

If you're curious i got the second part by using θ = (ω2 + ω1) t/2

So
θ = (60 + 10) 10/2