Linear and Rotational Kinetic Energy

[woaini]

Homework Statement

A solid cylinder is rolling without slipping. What fraction of its kinetic energy is linear?

Homework Equations

Ke=[itex]\frac{mv^2}{2}[/itex]+[itex]\frac{I(v/r)^2}{2}[/itex]

The Attempt at a Solution

Ke=[itex]\frac{mv^2}{2}[/itex]+[itex]\frac{(v/r)^2}{2}[/itex]*[itex]\frac{(mr)^2}{2}[/itex]

Ke=[itex]\frac{3(mv)^2}{4}[/itex]

Linear Ke =[itex]\frac{(mv)^2}{4}[/itex]

Fraction of Linear Ke = [itex]\frac{linear Ke}{Ke}[/itex] = [itex]\frac{(1/4)}{(3/4)}[/itex] = [itex]\frac{1}{3}[/itex]

 

[tiny-tim]

hi woaini!

Linear Ke =[itex]\frac{(mv)^2}{4}[/itex]

uhh?

 

[haruspex]

Ke= ... *[itex]\frac{(mr)^2}{2}[/itex]

The m should not be getting squared. You need to be more careful with the parentheses.

 

[woaini]

hi woaini!


uhh?

Nvm it should be 1/2. Therefore I should get a final answer of 2/3 linear kinetic energy?

 

[tiny-tim]

Therefore I should get a final answer of 2/3 linear kinetic energy?

yup!

btw, a way of checking this is to say the mass is m, the "rolling mass" (= I/r²) = m/2, so the "effective mass" is 3m/2, and eg the acceleration down a slope must be multiplied by 2/3 (but don't use that in the exam )

(for a sphere, it's m + 2m/5 = 7m/5, and the acceleration must be multiplied by 5/7)