Flywheel with kinetic friction

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Homework Help Overview

The problem involves determining the average magnitude of the couple due to kinetic friction in a flywheel that coasts to rest from a specified angular velocity. The context includes concepts from rotational dynamics and friction.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the conversion of weight to mass and the appropriate use of units in the Imperial system. There is an exploration of the calculations related to kinetic energy and work done by friction.

Discussion Status

Some participants have provided guidance on unit conversions and the significance of using the correct units for mass and weight. There is acknowledgment of a potential error in the original poster's calculations, particularly regarding the gravitational constant used.

Contextual Notes

Participants note the importance of using the correct gravitational constant for the Imperial system, highlighting the difference between using 9.81 m/s² and 32.2 ft/s². There is an emphasis on ensuring clarity in distinguishing between weight and mass in the calculations.

JJBladester
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Homework Statement


It is known that 1500 revolutions are required for the 6000-lb flywheel to coast to rest from an angular velocity of 3000 rpm. Knowing that the radius of gyration of the flywheel is 36 in., determine the average magnitude of the couple due to kinetic friction in the bearings.

Answer:

\left |M \right |=87.8lb\cdot ft

Homework Equations



KE_1+Work=KE_2

Work_{1\rightarrow 2}=M\theta

The Attempt at a Solution



It takes 1500 revolutions to come to rest and 1 revolution = 2\pi, so 1500 revolutions = 3000\pi.

W = 600lbs

m=\frac{W}{g}=\frac{6000}{9.81}=612lbs

\omega _1=300\frac{rev}{min}*\frac{2\pi}{1rev}*\frac{1min}{60sec}=10\pi\frac{rad}{sec}

Radius of gyration (k) = 36 in = 3 ft

KE_1=\frac{1}{2}\bar{I}\omega_1^{2}=\frac{1}{2}\left (k^2m \right )\omega_1^{2}=\frac{1}{2}\left (3^2*612 \right )\left (10\pi \right )^{2}=2.72e^6J

Work_{1\rightarrow 2}=M\theta=M(3000\pi)

KE_1+Work=0

2.72e^6=\left (-3000\pi \right )M

M=-288.6lb\cdot ft

I'm off... but where?
 
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When using weight in pounds, you must divide by g in units of ft/sec^2 in order to get the mass in slugs
 
PhanthomJay said:
When using weight in pounds, you must divide by g in units of ft/sec^2 in order to get the mass in slugs

It's also customary to write lbf for weight and lb for mass (slugs). Aside from units do you see where my method is incorrect? Thanks in advance Phantom. Happy Thanksgiving.
 
JJBladester said:
It's also customary to write lbf for weight and lb for mass (slugs). Aside from units do you see where my method is incorrect? Thanks in advance Phantom. Happy Thanksgiving.
Thank you, same to you.

Your method is fine, but when using the Imperial System of measure, always always always, in Physics, express force units in pounds and mass units in slugs. 1 slug weighs 32.2 pounds on Planet Earth (W = mg), and a force of 1 pound will give amass of 1 slug an acceleration of 1 m/s^2. Any other system of units, if youare not using SI, will lead you astray, guaranteed.
 
PhanthomJay said:
32.2 pounds

... Yeaaaaa, I used 9.81. Now the math works out. As a certain TV character would say, "DOH!"
 

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