Calculating Rotations and Linear Acceleration in a Flywheel System

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Homework Help Overview

The discussion revolves around a flywheel system, specifically focusing on calculating the number of rotations before coming to rest after steam is shut off, and determining the net linear acceleration of a particle at a given distance from the axis of rotation while the flywheel is in motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of angular motion formulas and their analogs to kinematics. There is an exploration of how to relate angular velocity and acceleration to linear quantities. Questions arise about unit conversions and the correct application of formulas.

Discussion Status

Some participants have made progress on the first part of the problem, while others are working through the second part and seeking clarification on their calculations. There is an ongoing exchange of ideas regarding the correct approach and necessary conversions, with no explicit consensus reached yet.

Contextual Notes

Participants mention missing information from prior lessons and express uncertainty about unit conversions, particularly converting revolutions per minute to radians per second.

PhysicsPhun
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I missed a week of school so I'm a bit behind, i do have a book so i need just a bit of guidance. Any extra help would be appreciated.

Question Part 1
The flywheel of a steam engine runs with a constant angular speed of 115 rev/min. When steam is shut off, the friction of the bearings and the air (assumed constant) brings the wheel to rest in 2.20 hours. How many rotations does the wheel make before coming to rest?

Part 2
Consider a particle that is located at a distance of 37.0 centimeters from the axis of rotation when the flywheel is turning at 0.390 rev/min. What is the magnitude of the net linear acceleration of this particle?
 
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Angular motion formulas have a direct analog to kinematics formulas. The formula relating velocity and acceleration in kinematics,

v = at

simply becomes

[tex]\Delta \omega = \alpha \Delta t[/tex]

where omega is the angular velocity and alpha is the angular acceleration. From that formula, you can calculate the acceleration. Then we have the equation

[tex]\Delta(v^2) = 2a\Delta x[/tex]

becomes

[tex]\Delta(\omega^2) = 2\alpha \Delta \theta[/tex]

from which you can calculate theta.

For the second part, centripetal acceleration is related to linear velocity by

[tex]a = \frac{v^2}{r}[/tex]

and linear velocity is related to angular velocity by

[tex]v= \omega r[/tex]

Combine the two and you're done.

cookiemonster
 
Got the first part, and now workin to understand the second. I'll let you know. Thanks.
 
For Part two, this is what I've come up with and I'm stuck on.
w = 115 rev/min
r = .37 m

Does v^2 in the first equation equal (wr)^2?

I missed all the work on Omega and such, I'm having a lot of trouble with some work.
 
Yes. v^2 = (wr)^2.

cookiemonster
 
Therefore..
(wr)^2/.37 should give me my answer?
 
Yes.

cookiemonster
 
I'm getting 4893.25.. m/s^2 and that's not right.

Should i be converting 115 rev/min to something?
 
PhysicsPhun said:
I'm getting 4893.25.. m/s^2 and that's not right.

Should i be converting 115 rev/min to something?

radians/min?
 
  • #10
I'll look up how to do that. or does anyone know off hand?
 
  • #11
One revolution = 2π radians;
One minute = 60 seconds.
 
  • #12
Didn't you say that it was going .390 rev/min?

(.390rev/min)(1min/60sec)(2pi rad/1rev) = .0408 rad/s

cookiemonster
 
  • #13
Consider a particle that is located at a distance of 37.0 centimeters from the axis of rotation when the flywheel is turning at 0.390 rev/min. What is the magnitude of the net linear acceleration of this particle?

Ya you're right Cookie.. I was lookin at the first part.
(wr)^2/r
(.0408*.37)^2/.37

a =.0006159 m/s^2.. Does that look right?
 
  • #14
.39 rev/min isn't very fast.

cookiemonster
 
  • #15
so my answer is on target?
 

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