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Kinetic energy of a rolling sphere

  • Thread starter kayron
  • Start date
10
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1. Homework Statement

A sphere of mass 50gm and radius 10cm rolls without slipping with a velocity of 5cm/s.
Its total kinetic energy in ergs is?
 

rl.bhat

Homework Helper
4,433
5
Total kinetic energy = linear KE + rotational KE.
 
10
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ok i got the answer. thank you.

i have a question though, i wanna know what the question means when it says that the sphere rolls without slipping??
 

jhae2.718

Gold Member
1,110
20
When an object rolls without slipping, it means the velocity of the center of mass is equal to the radius times angular velocity, [itex]v_{cm}=r\omega[/itex]

This is called the nonslip condition. When an object rolls with slipping, the linear velocity is not [itex]r\omega[/itex]
 

gneill

Mentor
20,488
2,610
ok i got the answer. thank you.

i have a question though, i wanna know what the question means when it says that the sphere rolls without slipping??
It means that the sphere where it meets the surface it's rolling on, does not slide -- the instantaneous point of contact is stationary with respect to the "ground" surface.
 
10
0
When an object rolls without slipping, it means the velocity of the center of mass is equal to the radius times angular velocity, [itex]v_{cm}=r\omega[/itex]

This is called the nonslip condition. When an object rolls with slipping, the linear velocity is not [itex]r\omega[/itex]
okay, then what would it be?
 

jhae2.718

Gold Member
1,110
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It's situation dependent. An example of a slipping condition would be the tires on your car skidding, where the wheels would both rotate and translate forward, so the [itex]r\omega[/itex] term would be less than v at the CM.
 
10
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how do you know it would be less??
 

jhae2.718

Gold Member
1,110
20
With that example I just made the assumption that wheels were slipping forwards; then both the rotational and translational terms would contribute to the velocity at the center of mass.

It really depends on the situation, though.
 
10
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okay
 

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