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Homework Help: Object Rolling up ramp

  1. Jul 8, 2007 #1
    1. The problem statement, all variables and given/known data

    A ring of mass 2.83 kg, inner radius 5.30 cm, and outer radius 6.95 cm rolls (without slipping) up an inclined plane that makes an angle of θ=37.4°, as shown in the figure below.
    http://capa.physics.mcmaster.ca/figures/sb/Graph11/sb-pic1106.png [Broken]

    At the moment the ring is at position x = 2.18 m up the plane, its speed is 2.69 m/s. The ring continues up the plane for some additional distance and then rolls back down. It does not roll off the top end. How much further up the plane does it go?

    2. Relevant equations
    I think this has something to do with Kinetic energy of the rolling object...

    KE=1/2Iw^2+1/2mv^2 (where I= MR^2)

    w=2pi rad/perod

    delta xcm= vcm delta t (to find the period)


    sin 37.4=opp/hyp (which I put as the distance x)

    3. The attempt at a solution

    I first calculated for the period using

    delta xcm= vcm delta t (to find the period)
    2.18=2.69 t

    w=2pi rad/period
    =2pi rad/0.810s
    =7.76 rad/s


    Sin 37.4= opp/2.18m
    opp (height when ring is at x)= 1.32m

    mgh at this point= (2.83kg)(10N/m^2)(1.32m)

    Final PE when the v of the ring=0 is 37.56+11.52
    =48.88J at the highest point when all the KE would have transferred to PE


    To find the total length of the hypotentus at this height:

    =2.84m (final x)

    Distance that still needs to be travelled:
    2.84- 2.18 (initial x)= 0.663m

    But this answer is not right. I feel like that there's something that I missing or not relating properly! If anyone can provide some insight I'd really appreciate it! :)
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jul 8, 2007 #2


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    Staff Emeritus
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    Gold Member

    You are working to hard.

    YOu have the KE of the object. To find how much higher it goes set:

    KE = PE = mgh

    so [tex] h = \frac {KE} {mg} [/tex]

    and to get how much further up the ramp

    [tex] d = \frac h {sin \theta} [/tex]
  4. Jul 8, 2007 #3

    Doc Al

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    Staff: Mentor

    Nothing wrong with using energy methods to solve this. But realize, since they gave inner and outer radii, that this is not a thin ring. So you'll need to revise that formula for I.

    You can solve for the angular speed using the condition for rolling without slipping: [itex]v = \omega r[/itex].

    Not sure what you are doing here. The 2.18 m is irrelevant to this part of the problem. And the speed is not constant as the hoop rolls up the incline.
  5. Jul 8, 2007 #4
    Thank you for your replies!!!

    What both of you are saying makes sense, finding angular speed is definately much easier with v=wr, but which radius would I use? the inner, outer or a total of both?

    I tried to solve the problem again using the approaches that you suggested but i'm still not getting the right answer.

    I tried answering the question first by finding angular speed by using r=0.1225m which is the total of inner and outer radii, the final answer that I got didn't work, and then I used the radii of only the outer portion to find angular speed and the final answer that I got for that didnt' work either.

    Now Doc Al, you were saying that "I" should be adjusted because the ring is not a thin ring, but how do I go about adjusting that? I got MR^2 from a table in my textbook, and that's where our prof told us to look if we ever needed to find a value for "I". And also when I replace the R value for the MR^2 of I, would I use the total radius of the object including the inner portion or should i use either inner or outer??

    Thanks so much!!!
  6. Jul 8, 2007 #5

    Doc Al

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    Staff: Mentor

    Which radius do you think relates translational to rotational speed? Hint: If the inner radius changed, would it make a difference?

    Adding the radii makes no sense!

    Just like you looked up the formula for a thin cylinder or ring, look up the formula for a thick cylinder. Hint: It involves both radii--but don't just add them.
  7. Jul 8, 2007 #6
    I ran out of tries and I still didn't get the question right.. :(

    In this case, would the outer radius matter more because that's where most of the mass is concentrated? I really had trouble thinking about this because if you decrease the inner radius (ie. bring more of the mass to concentrate at the center, it will decrease the moment of inertia) but at the same time because the object is a cylinder, wouldn't the mass concentration around the outside be more important and offer the drive for the translational and rotational motion?

    Also I couldn't find a value for a thick ring, the only thing that came close to it was a cylinder of 1/2MR^2, is this the one that I should have used?

    This question really stumped me, actually still does!

    But thanks for your help anyways! :)
  8. Jul 8, 2007 #7

    Doc Al

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    Staff: Mentor

    Sorry to hear that.

    There are two different issues:
    (1) For the rolling without slipping condition ([itex]v = \omega r[/itex]), only the outer radius matters--that's the one doing the rolling.
    (2) For the moment of inertia calculation, both radii matter. And yes, as the inner radius decreases, the moment of inertia also decreases. (Good thinking.)

    No, that's for a solid cylinder. Look up the one you need here: http://en.wikipedia.org/wiki/List_of_moments_of_inertia" [Broken]
    Last edited by a moderator: May 3, 2017
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