Linear and Rotational Kinetic Energy

  • Thread starter woaini
  • Start date
  • #1
58
0

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]
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
hi woaini! :smile:
Linear Ke =[itex]\frac{(mv)^2}{4}[/itex]

uhh? :confused:
 
  • #3
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,203
6,814
Ke= ... *[itex]\frac{(mr)^2}{2}[/itex]
The m should not be getting squared. You need to be more careful with the parentheses.
 
  • #4
58
0
hi woaini! :smile:


uhh? :confused:

Nvm it should be 1/2. Therefore I should get a final answer of 2/3 linear kinetic energy?
 
  • #5
tiny-tim
Science Advisor
Homework Helper
25,832
251
Therefore I should get a final answer of 2/3 linear kinetic energy?

yup! :biggrin:

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 :wink:)

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

Related Threads on Linear and Rotational Kinetic Energy

  • Last Post
Replies
8
Views
6K
Replies
4
Views
3K
  • Last Post
Replies
3
Views
1K
Replies
4
Views
2K
Replies
2
Views
2K
Replies
1
Views
2K
Replies
15
Views
875
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
6K
Top