Work and Rotational Kinetic Energy

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SUMMARY

The discussion focuses on calculating the work required to stop a rotating wheel, specifically a 38.0 kg thin hoop with a radius of 1.13 m, rotating at 280 revolutions per minute (rev/min). The relevant equation for change in kinetic energy is established as ΔKE = (1/2)Iωi² - (1/2)Iωf², where I is the moment of inertia calculated as I = (1/2)mr². The initial kinetic energy is computed, and it is clarified that time does not factor into the work calculation, leading to a resolution of the problem.

PREREQUISITES
  • Understanding of rotational dynamics and kinetic energy
  • Familiarity with the moment of inertia formula for thin hoops
  • Knowledge of angular velocity conversion from revolutions per minute to radians per second
  • Basic algebra for manipulating equations
NEXT STEPS
  • Learn about the relationship between linear and angular motion
  • Study the concept of work-energy theorem in rotational systems
  • Explore the calculation of moment of inertia for different shapes
  • Investigate the effects of friction and torque on rotational motion
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Physics students, educators, and anyone interested in understanding the principles of rotational kinetic energy and work calculations in mechanical systems.

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Homework Statement


A 38.0 kg wheel, essentially a thin hoop with radius 1.13 m, is rotating at 280 rev/min. It must be brought to a stop in 15.0 s.
(a) How much work must be done to stop it?

Homework Equations


change in KE = (1/2)Iwi^2 - (1/2)Iwf^2
I = (1/2)mr^2

The Attempt at a Solution


(1/2)[(1/2)(38)(1.13)^2](28pi/3) = 355.686 kg(m)^2/s

I don't see where the time comes into play though..can anyone help?
 
Last edited:
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(1/2)[(1/2)(38)(1.13)^2](28pi/3)
It should be
(1/2)[(1/2)(38)(1.13)^2](28pi/3)^2

To find the work done, time is not necessary.
 
Oh ok, I get it now!
Thank you so much!
 

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