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Another few questions from klppner's mechbook.

  1. Apr 15, 2007 #1
    i hope the fact that the next questions are from kleppner's dont make people to be detered from helping me. (-:
    (the attached file has the figures of the next questions.
    the questions:
    A solid rubber wheel of radius R and mass M rotates with angular velocity w0 about a frictionless pivot. a second rubber wheel of radius r and mass m, also mounted on a frictionless pivot, is brought into contact with it.
    what is the final angular velocity of the first wheel?
    A cone of height h and base radius R is free to rotate about a fixed vertical axis. It has a thin groove cut in the surface. The cone is set rotating freely with angular speed w0, and a small block of mass m is released in the top of the frcitionless groove and allowed to slide under gravity. Assume that the block stays in the groove. Take the moment of inertia of the cone about the vertical axis to be I0.
    a. what is the angular velocity of the cone when the block reaches the bottom?
    b. find the speed of the block in inertial space when it reaches the bottom?

    here are my answers
    the momet of inertia at the point of contact is the sum of the contribution from the rubber wheel with radius R and with the contribution of the wheel of r, according to steiner's thoerem: I=I_cm+Md^2 where d is the distance of I_cm from the point of contact so we have: [tex]I_{R}=MR^2+MR^2=2MR^2[/tex]
    and [tex]I_{r}=Mr^2+Mr^2=2Mr^2[/tex]
    here by conservation of angular momentum the torques are getting cancelled at the point of contact.
    so we have [tex](I_{R}+I_{R})\omega=MR^2\omega_0[/tex]
    is this correct?
    basically i think this is a question on energies:
    we have the first enrgy equals mgh and we it gets at the bottom it gains rotational energy of (1/2)I_0w_0^2 and the mv^2/2 where v=wr where w is the ang velocity of the cone when the block reaches the bottom of the cone.
    if this is correct then the answer to b, is sqrt(v^2+2gh).
    am i on the right track?

    Attached Files:

  2. jcsd
  3. Apr 15, 2007 #2
    lqg, while we are waiting for the attachment to clear, how do you like Kleppners text? I'm looking to buy a good intro/inter mechanics text as have nothing currently but memories and the net and I hear K&K treats at deeper level than all the rest. Certainly problems above are more challenging than the average posted.
  4. Apr 15, 2007 #3
    re first problem, the second wheel is smaller from sketch and implied to be of both different mass and size by lower case r and m. It also looks like the axes of rotation for the two will line up, so no need to consider parallel offset. But the final I is not two MR^2 as you suggest.

    Re second problem, I now know I need to own this book. You answered my question indirectly; unfortunately I can't answer yours. Frankly I cannot even read all of the question, but even if your supposition was right, you haven't answered the bear of the question which is what is the final angular velocity? Surely the total system Ir grows as the mass slides from 0 to r, and as it contributes nothing towards tangential accelaration, I'm sort of at a loss wondering where the mgh goes? Fabulous problem!
  5. Apr 16, 2007 #4
    can you explain why it isn't 2MR^2, MR^2 is I_cm of the wheel with R radius, but we want the moment of inertia in the place of contact so we need to use steiner's theorem.
  6. Apr 16, 2007 #5
    why can't you read all of the second question?
    i posted it in my opening post question number 6.33.
    my attached file is mainly for the sketches.
    about the book, im not sure this is the best book, perhaps it's best that you first lend it from your library, afterwards if it suits you purchase it.
  7. Apr 16, 2007 #6
    Maybe. when I looked at this problem, I thought it was like a collision problem in translational motion where two blocks collide, one initially at rest, and then the two stick together: the final velocities at their respective rims have to be equal (the blocks stuck together) and momentum conserved.

    The final combined I of the systems would just be the sum of the two I's rotating about their own centers. But it looks as if both mass,m and size,r of the second wheel are different.
    leading to something like,
  8. Apr 16, 2007 #7
    Yes, you're right I can make out sketch and read problem. Wasn't sure quote of problem was exact, but it seems so.

    Just not sure that treating the problem as simply one where I, the systems total moment of inertia varies with position of block.

    If it were flat disk with same groove and m in center moving radially outward, this approach should be fruitful. What are your thoughts on the first part?
  9. Apr 16, 2007 #8
    but shouldnt you calculate the moment of inertia at the point of contact?
    that's really the thing that makes it tough, i thought i should calculate MI at the point of contact and thus using steiner's theorem.
    and i have questions when im not sure when to calculate with regard to steiner's thoerem or to disregard it, obviously when the motion is around a specific point i should calculate the moment of inertia of the body perpendicular to the point which around it there's the motion.
    do you have any tips when to use it?
  10. Apr 16, 2007 #9
    My understanding, which I must warn you is quite incomplete, is that Steiner's theorum is used primarily for computing moments of inertia when rotation occurs thru an axis other than thru the Center of mass, which is normally how I's are computed, so if you look up I for a disc, or sphere, or whatever, you need to make adjustment if axis is not thru Cm.

    However the moment of inertia can be computed around any axis if you want to solve from original principles. Steiner's just makes it easy to just lookup and adjust accordingly, versus doing the double integration required if doing from scratch. So if you have a square sheet and punch a hole thru center, I is just what one would look up. Now if you punched hole out near an edge you need to adjust along one or both dimensions.

    Now point of contact of force is important, as torgue is cross product, but the I used depends only on axis of rotation. Not sure if I'm helping at all here....
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