1. Oct 14, 2007

### BuBbLeS01

1. The problem statement, all variables and given/known data
The flywheel of a steam engine runs with a constant angular speed of 146 rev/min. When steam is shut off, the friction of the bearings and the air brings the wheel to rest in 1.2 h.

A.) What is the magnitude of the constant angular acceleration of the wheel in rev/min2?

B.) How many rotations does the wheel make before coming to rest?

C.) What is the magnitude of the tangential component of the linear acceleration of a particle that is located at a distance of 60 cm from the axis of rotation when the flywheel is turning at 73.0 rev/min?

D.) What is the magnitude of the net linear acceleration of the particle in the above question?

2. Relevant equations
Alpha = Change in W/Change in T
At = alpha * r
Theta = Theta initial + W initial + (alpha * T)/2
V = WR
A = (Vf - Vo)/T

3. The attempt at a solution
This is a problem we did on a previous homework but I am trying to do it again to study for my test tomorrow but I am not understanding why I am doing what.

A.) Alpha = Change in W/Change in T
(146rev/min)/72 min = 2.03 rev/min^2

I though that W had to be in rad not rev for the equation to work. I know it asks for it in rev/min^2 so W can be in rad/min and rev/min???

B.) Theta = Theta initial + W initial - (alpha * T)/2
Why do I change it to - alpha? Is it because it is slowing down?

146 rev/min * 72 min - (2.03 rev/min^2 * 72^2)/2 = Theta
So we use W as rev/min not radians because thats what we used previously?

C.) At = Alpha * r
(2.03 * 2pi/3600) * .60 = .21 cm/s^2
Now why are we dividing by 3600 and not just 60? I thought we were trying to get it to seconds so why is not just 60? Why are we NOW changing alpha to rad??? Why can't we use rev like we have been? I am getting myself so confused now!

D.) I haven't started to look over this one yet I just really need help with A-C

Thanks so much!!!

2. Oct 14, 2007

### Staff: Mentor

In this equation, $\omega$ can have any units you like. It's only when relating angular and linear (tangential) quantities as in $v = \omega r$ that $\omega$ has to be in radians/second.

Yes, $\alpha$ is negative since its slowing down. Again, you can use any units you like as long as you're consistent.

That equation should be:
$$\theta_f = \theta_i + \omega_i t + 1/2 \alpha t^2$$
($\alpha$ is negative in this case.)

Now this is a situation where you need to use rad/sec, not rev/min. And since the angular acceleration has units of $1/{min}^2$, you need to divide by $(60 s)^2 = 3600 s^2$

Also, since you use a radius of 0.6 m (= 60 cm), your answer will have units of m/s^2, not cm/s^2.

3. Oct 14, 2007

### BuBbLeS01

so tangential acceleration is a linear quantity?

4. Oct 14, 2007

### Staff: Mentor

Absolutely--it's measure in m/s^2 like any other linear acceleration. (Read this: Rotational-Linear Parallels)

5. Oct 14, 2007

### BuBbLeS01

Thank you so much, that page will be very helpful!

So for part D I was trying to use the following equations...
v = wr
a = vf-vi/t

but you can't cause it's not the right answer. Why can't these equations be used since it asks for net linear acceleration?

I am supposed to use Ac = v^2/r
So Ac is considered net linear acceleration always?

6. Oct 14, 2007

### Staff: Mentor

The linear acceleration has two components: a tangential component (which you figured out in part C) and a centripetal component. Figure out the centripetal component and then find the magnitude of the total (net) acceleration.

The centripetal acceleration will only equal the net acceleration when something rotates at constant angular speed--in that case there's no tangential acceleration to worry about.

7. Oct 14, 2007

### BuBbLeS01

So it just like finding the resultant vector of At and Ac...
At = 0.00212 m/s^2
Ac = 35.255 m/s^2
sqrt 0.00212^2 + 35.255^2 then multiply by 100 because I need the answer in cm so I get 3525 cm/s^2

8. Oct 14, 2007

### Staff: Mentor

That's the idea. (Recheck your calculation of At--I get a different value.)

9. Oct 14, 2007

### BuBbLeS01

At = (2.03*2pi/3600) * .60 = 0.0021258

Last edited: Oct 14, 2007
10. Oct 14, 2007

### BuBbLeS01

I keep getting the same 0.00212 answer

11. Oct 14, 2007

### Staff: Mentor

2.03 is the angular acceleration in rev/min^2. What would it be in rad/s^2?

12. Oct 14, 2007

### BuBbLeS01

but thats why I multiplied it by 2pi and divided by 3600

13. Oct 14, 2007

### Staff: Mentor

My bad--you are correct. (I think you had a typo in your original version of post #9 and I wasn't paying close enough attention.)

14. Oct 14, 2007

### BuBbLeS01

Yea I did I am sorry I edited a couple minute later because I forgot the /3600 part. Thanks for you help!