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Homework Help: Rotational Motion Problem Help

  1. Oct 31, 2012 #1


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    A solid, uniform disk of radius 0.250 m and mass 57.7 kg rolls down a ramp of length 3.70 m that makes an angle of 14° with the horizontal. The disk starts from rest from the top of the ramp.

    (a) Find the speed of the disk's center of mass when it reaches the bottom of the ramp.

    (b) Find the angular speed of the disk at the bottom of the ramp.

    I've tried the first part.

    I set the equation up as...

    mgh=1/2mv2+1/2(1/2) I ω^2



    Might have been wrong there on the second part of the equation? Does the Iw^2 just turn into mv^2 since its not slipping?

    I then added together the fractions and got


    Plugged in the values but didnt get the right answer. About 99% sure I went somewhere wrong in getting the right equation. I noticed too I never used the radius or angle, not sure if theyre needed. Any help would be great thanks.
  2. jcsd
  3. Oct 31, 2012 #2
    2 does not equal mv2

    what's I for a solid uniform disk and what's the relationship between ω and v?
  4. Oct 31, 2012 #3


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    Solid uniform disk is I=1/2mr^2.
  5. Oct 31, 2012 #4
    and omega is?
  6. Oct 31, 2012 #5


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    angular velocity and v is just linear, normal velocity. They're both unknowns in the equation and you can't have two unknowns so I wasn't sure what to do with omega. But I know omega= v/r
  7. Oct 31, 2012 #6
    well, if you convert omega into v then you only have to deal with v. It condenses your equation and makes things easier to deal with.

    it's also okay that v is an unknown, because we are applying conservation of energy.

    write out the initial energy of the disk and the final energy, and set them equal to each other

    then apply the constraints of the problem, which are: initial velocity is zero, and also when the disk is at the bottom of the ramp we can say that the height is equal to zero, since we are only worried about the *change* in potential energy

    EDIT: oops sorry if I sounded like I thought you didn't know how to do conservation of energy, I was just being kind of mechanical >.>
    Last edited: Oct 31, 2012
  8. Oct 31, 2012 #7
    anyways, your mistake was just that you converted the angular kinetic energy improperly

    also the angle of the hill doesn't matter since it's conservation of energy, which only concerns itself with initial and final conditions
  9. Oct 31, 2012 #8


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    ok your post was kind of confusing but I think I understand.

    so would it still be gravitational potential energy = KE rotational + KE translational?
  10. Oct 31, 2012 #9
    well so the total amount of energy at the beginning needs to be equal to the amount of energy at the end


    PEi + KErot,i + KEtrans,i = PEf + KErot,f + KEtrans,f

    where the little i means "initial" and the little f means "final". PE is potential energy, mgh.

    but KErot,i and KEtrans,i are both zero and PEf is zero

    so we end up with what you just said

    PEi = KErot,f + KEtrans,f
  11. Oct 31, 2012 #10


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    Yeah i just didnt include the "i's" and "f's" for initial and final. ok so I have mass, g is 9.8, and h would be length of the ramp, KE=1/2mv^2 for translational KE, and then rotational KE is where I'm having trouble. I for a solid disk is I=1/2mr^2. So would the rotational KE be 1/4mv^2 ? Edit-nevermind thats wrong thats what I originally thought it was. Not sure what to do about the rot KE.
  12. Oct 31, 2012 #11
    oh, well also it was wrong of me to say the angle of the hill doesn't matter, since we want to use the *height* difference. Since we only know the *length* of the ramp, we need to use some trigonometry to find the height of the ramp.

    So just take 1/2Iω2 and substitute in your expressions for I and ω
  13. Oct 31, 2012 #12


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    How would I include the angle. Would I do like sin14= opp/3.7

    3.7 m was the length of the ramp. then I just solve for the opposite to get the height of the ramp? and thatd be h.

    and by substituting the expressions you mean putting in mr^2 for I and v/r for ω?

    Sorry this is taking me so long to understand and figure out. This class has nothing to do with my major. Just one of those classes that has to be taken. So confusing.
  14. Oct 31, 2012 #13
    oh don't worry about it

    and yeah you did the trig right. It always helps me to draw the triangle so I can visualize it better.

    And yes, you just want to replace ω with v/r and replace I with (1/2)mr2.
  15. Oct 31, 2012 #14


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    Ok, now the only problem I have is I dont know what to do since I have two unknowns. "v" shows up twice in the equation. to find v would I come up with some equation like v= square root and then a bunch of stuff. 2mg over something. i dont know.
  16. Oct 31, 2012 #15
    well the vs are the same thing

    This is because the angular velocity is *directly related* to the translational velocity of the object. If you know the radius, you can go between the two.

    so you should have

    [itex]mgh = \frac{1}{2}mv^{2} + \frac{1}{2} \frac{1}{2}mr^{2}\frac{v^{2}}{r^{2}}[/itex]

    after substituting, right?

    a little more simplification gets us

    [itex]mgh = \frac{1}{2}mv^{2} + \frac{1}{4}mv^{2}[/itex]

    and then just solve for v from there.
  17. Oct 31, 2012 #16


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    Ok wow. Yeah I was just not realizing these simple things. I was making this much harder than it is. I plugged everything in and got the right answer. Thank you so much for being patient with me and helping me on this.
  18. Oct 31, 2012 #17
    oh it's no problem

    with conservation of energy problems the best strategy is to take things slow and be very methodical with your work
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