Rotational Kinetic energy of bar, and new kinetic energy

  • #1
DracoMalfoy
88
4

Homework Statement


A bar of length 2.5m and mass 5kg, whose rotation point is at its center, rotates at 5 rad/s. What is the rotational kinetic energy of the bar?


If a point mass of mass 1.5kg is added to each end of the bar, assuming the angluar velocity is the same, what is the new kinetic energy?

a. 15.6J, 78.9J

b. 32.6J, 91.1J

c. 32.6J, 267J

d. 15.6J, 35.8J

e. 10.7J, 42.9J

Homework Equations


KE=.5⋅ I⋅ω^2

I=m⋅r^2

The Attempt at a Solution


[/B]
Im not sure where to start on this. I really need help. I've been working on this for 20 minutes.

ω= 5rad/s
r=2.5m= 1.25m
m=5kg
 

Answers and Replies

  • #2
haruspex
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Homework Statement


A bar of length 2.5m and mass 5kg, whose rotation point is at its center, rotates at 5 rad/s. What is the rotational kinetic energy of the bar?


If a point mass of mass 1.5kg is added to each end of the bar, assuming the angluar velocity is the same, what is the new kinetic energy?

a. 15.6J, 78.9J

b. 32.6J, 91.1J

c. 32.6J, 267J

d. 15.6J, 35.8J

e. 10.7J, 42.9J

Homework Equations


KE=.5⋅ I⋅ω^2

I=m⋅r^2

The Attempt at a Solution


[/B]
Im not sure where to start on this. I really need help. I've been working on this for 20 minutes.

ω= 5rad/s
r=2.5m= 1.25m
m=5kg
The equation you quote for I is only for a point mass about another point (or a ring, or hollow cylinder, about its mass centre).
You need a different formula for a thin rod about its mass centre. Nothing in your notes?
 
  • #3
DracoMalfoy
88
4
thin rod about its mass centre

1/12⋅m⋅L^2?
 
  • #5
DracoMalfoy
88
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I did that and got 25.5. do i sub that into the other equation?
 
  • #6
haruspex
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I did that and got 25.5. do i sub that into the other equation?
That's rather inaccurate. Please post your working.
 
  • #7
DracoMalfoy
88
4
That's rather inaccurate. Please post your working.

I=(1/12)(5⋅9.81)(2.5)^2

I guess i shouldn't multiply the 5 by 9.8? because I am getting 32.6 as the answer when i sub it into the other equation when i do it the other way
 
  • #8
haruspex
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i shouldn't multiply the 5 by 9.8?
Why would you multiply by g? The formula references the mass, m, not the weight. The moment of inertia would be the same on the moon as on Earth.
 
  • #9
DracoMalfoy
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I guess I'm so used to doing it since we were doing that all through the last unit. Do i add the point masses to the kinetic energy?
 
  • #10
haruspex
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Do i add the point masses to the kinetic energy?
You add the contribution from the point masses, yes.
 
  • #11
DracoMalfoy
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But I am getting 52.2 as the answer when i do that .-.

Ipm=m⋅r^2 right? and the radius is 1.25. i multiplied by the mass.
 
  • #12
haruspex
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But I am getting 52.2 as the answer when i do that .-.

Ipm=m⋅r^2 right? and the radius is 1.25. i multiplied by the mass.
I cannot see how you get 52.2, whether you mean for one mass, both masses, or the whole system. Please post your working.
 
  • #13
DracoMalfoy
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I used the point mass formula for both of the masses ._. then added to the kinetic energy
 
  • #14
gneill
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I used the point mass formula for both of the masses ._. then added to the kinetic energy
Please show your calculation explicitly. Type them out so that they can be examined and commented upon.
 
  • #15
DracoMalfoy
88
4
I=1/12⋅m⋅L^2
I= 1/12⋅5⋅2.5^2
I=2.604

KE=1/2⋅I⋅ω^2
KE=1/2⋅2.604⋅5^2
KE= 32.6

For point masses

I=1/3⋅1.5⋅1.25^2
I=.7813

KE=1/2⋅.7813⋅5^2
KE= 10

Then i tried to add 20 to the 32.6 and I am not getting the answer. I am probably doing this completely wrong.
 
  • #16
haruspex
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For point masses

I=1/3⋅1.5⋅1.25^2
Where is the 1/3 coming from? They are point masses. Use the formula for that.
 

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