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Uniform Hoop Hit By a Small Sticky Ball of Putty

  1. Jul 30, 2005 #1
    I had this question on my physics final this summer and I can't figure out the answer. A uniform hoop of mass M and radius R is hanging from a frictionless pivot point. A sticky putty ball is thrown horizontally at the hoop with linear momentum "mVo", where m is the mass of the putty and "Vo" is initial velocity. The putty sticks to the hoop in such a way that a line drawn from the putty to the center of the hoop is parallel to the horizontal.

    Find the angular speed of the hoop imidiately after impact.
    Find the minimum mass m of the putty ball so that the hoop is able to complete a full vertical "loop" around it's pivot point.

    Now I tried thinking about this problem and I know that since the putty ball sticks to the hoop, mechanical energy is not conserved. At the time of impact, angular momentum is not conserved in the putty-hoop system dbecause the pivot point exerts an outside force. Also, linear momentum is not conserved because the pivot exerts and outside force also. (please tell me if my thinking is incorrect with any of these)

    From there, I'm stuck, because I dont know what conservation laws to apply and I also dont know how linear impulse of angular-impulse would help me either, but one of these methods MUST work (I hope)

    Here is a link to an image of the problem.
    http://img316.imageshack.us/img316/5499/hooputty9am.jpg [Broken]
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jul 30, 2005 #2

    Andrew Mason

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    This is just a conservation of linear momentum problem. But I am a little confused about what you mean by a vertical loop around the pivot point. Do you have a diagram?

  4. Jul 31, 2005 #3
    here's a link to the image. By vertical loop, I mean the hoop is able to complete a revolotion around it's pivot point (as if you pushed the hoola hoop hanging from your finger so that it went all the way around). The faded loops just demonstrate the motion that the second part of the question describes.

    http://img316.imageshack.us/img316/5499/hooputty9am.jpg [Broken]
    Last edited by a moderator: May 2, 2017
  5. Jul 31, 2005 #4


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    The impect is not elastic that is why we have to use conservation of angular momentum about the axis of rotaqtion (Pivot). The putty mooving horizontally along the horizontal radius is having an angular momentum of mV0*R about the axis of rotaqtion (Pivot).
    Conserve angular momentum about the pivot and then conserve kinetic energy after impect.
  6. Jul 31, 2005 #5


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    You can conserve angular momentum of the system (ie, the putty and the hoop) befroe and after impact about the pivot point because the torque due to external forces around the pivot point is zero.

    Once the putty collides with the hoop, it sticks to it. Therefore, both the putty and the hoop will travel with the same angular velocity. You can calculate this angular velocity by using conservation of angular momentum about the pivot point.

    Note that energy will not be conserved when the putty strikes the hoop because the collision is inelastic. But, after the putty sticks to the hoop and travels with a common angular velocity, energy will be conserved. Use this to calculate the answer to the second part of the question
  7. Jul 31, 2005 #6
    Why is the torque about the pivot pointzero after the impact? Before the impact, gravity causes no torque because it acts through the center of mass of the hoop, and the line of acting of gravity passes through the pivot point (thus, no torque). When the putty immediately sicks to hoop, doesn't the center of mass of the system shift horizontally a little bit? If that is the case, then gravity immediately creates a torque at the time of impact because gravity no longer has a line of action through the pivot point, which is why I dont understand how angular momentum is conserved. I appreciate the help guys, I'm having trouble determining what's wrong with my reasoning though. You are saying what my physics teacher said, but I didn't understand him either. Here's a link to a picture of what I'm thinking.
    http://img28.imageshack.us/img28/5144/hooputty21wq.jpg [Broken]
    Last edited by a moderator: May 2, 2017
  8. Aug 1, 2005 #7


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    Whether this shifting is instantaneous? No time taken in this shifting? What is the position of putty then?

    Actually we consider this as momentary impact; otherwise you have forgotten the torque due to the weight of the putty as well.

    We consider that the putty moving horizontally has the angular momentum mv0R, which is suddenly transferred to the putty hoop system. In the subsequent motion the gravity is considered and in place of the torque we are considering the energy conservation, as it is easy to calculate.
  9. Aug 1, 2005 #8
    Thanks I understand it now.
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