Kinetic energy of a rolling hoop

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SUMMARY

The discussion centers on calculating the work required to stop a 120 kg hoop rolling at a speed of 0.240 m/s. The kinetic energy (KE) of the hoop is derived using the formula KE = 0.5 * m * v^2 + 0.5 * (mR^2)(v/R)^2, resulting in a total KE of 6.912 J. Participants emphasize the importance of matching significant digits to the given data and suggest including an explanation of the relationship between total kinetic energy and the work required to stop the hoop.

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Jrlinton
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Homework Statement


A 120 kg hoop rolls along a horizontal floor so that its center of mass has a speed of 0.240 m/s. How much work must be done on the hoop to stop it?

Homework Equations


I of hoop=MR^2

The Attempt at a Solution


KE=0.5*m*v^2+0.5*(mR^2)(v/R)^2
=0.5*120kg*.24^2m/s+0.5*(120kg*R^2)(.24m/s^2/R^2)
=6.912 J

.
 
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Your calculation looks good. You might want to trim the significant digits in your result to match the given data. If this is a "show your work" type question rather than a web based "enter the solution" assignment, you might want to include a note as to why the total KE is the same as the work required.
 

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