How do I solve problems involving rotational motion?

  • Thread starter BuBbLeS01
  • Start date
  • Tags
    Test
In summary, the conversation discusses a problem involving a steam engine flywheel with a constant angular speed of 146 rev/min and its gradual stop due to friction. The conversation covers the calculation of the constant angular acceleration, the number of rotations made before coming to rest, and the magnitude of the tangential and net linear accelerations of a particle located at a distance of 60 cm from the axis of rotation. The conversation also addresses the use of radians and revolutions in equations and the difference between linear and tangential acceleration.
  • #1
BuBbLeS01
602
0
OMG Test Tomorrow! Please help!

Homework Statement


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?


Homework 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


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 that's 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!
 
Physics news on Phys.org
  • #2
BuBbLeS01 said:
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?
In this equation, [itex]\omega[/itex] can have any units you like. It's only when relating angular and linear (tangential) quantities as in [itex]v = \omega r[/itex] that [itex]\omega[/itex] has to be in radians/second.

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 that's what we used previously?
Yes, [itex]\alpha[/itex] is negative since its slowing down. Again, you can use any units you like as long as you're consistent.

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

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

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
so tangential acceleration is a linear quantity?
 
  • #4
BuBbLeS01 said:
so tangential acceleration is a linear quantity?
Absolutely--it's measure in m/s^2 like any other linear acceleration. (Read this: Rotational-Linear Parallels)
 
  • #5
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
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
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
That's the idea. (Recheck your calculation of At--I get a different value.)
 
  • #9
At = (2.03*2pi/3600) * .60 = 0.0021258
 
Last edited:
  • #10
I keep getting the same 0.00212 answer
 
  • #11
2.03 is the angular acceleration in rev/min^2. What would it be in rad/s^2?
 
  • #12
but that's why I multiplied it by 2pi and divided by 3600
 
  • #13
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
Yea I did I am sorry I edited a couple minute later because I forgot the /3600 part. Thanks for you help!
 

1. What is the OMG Test and why do I have to take it tomorrow?

The OMG Test, or "Oh My Goodness" Test, is a common phrase used to refer to any important or high-stakes exam. It is likely that you have to take it tomorrow because it is a scheduled exam for a particular course or subject.

2. How should I prepare for the OMG Test tomorrow?

The best way to prepare for any test is to review your notes, study materials, and practice problems or questions. Take breaks and get a good night's sleep to ensure you are well-rested for the test.

3. Will the OMG Test tomorrow cover everything we learned in class?

It is unlikely that the test will cover every single thing that was taught in class. However, it will likely cover the main topics and concepts that were discussed and emphasized by the teacher.

4. What happens if I don't do well on the OMG Test tomorrow?

If you don't do well on the test, it is important to not panic and instead reflect on what areas you struggled with. You can also talk to your teacher for tips on how to improve for future tests. One test does not define your overall understanding of the subject.

5. Can I use a calculator or notes during the OMG Test tomorrow?

This will depend on the specific test guidelines set by your teacher. Some tests may allow calculators or notes, while others may not. It's important to clarify this with your teacher beforehand to avoid any confusion during the test.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
21
Views
7K
  • Introductory Physics Homework Help
Replies
12
Views
2K
Replies
7
Views
235
  • Introductory Physics Homework Help
Replies
18
Views
3K
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
840
Back
Top