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Speed question

  1. Oct 23, 2006 #1
    Suppose in the figure attached View attachment Doc1.doc that the two hoops are rigidly fastened together. The system is released from the position shown and there is negligible friction between the large hoop and the table. How fast relative to the table is the center of the large hoop moving when the centers of the two hoops lie on a vertical line? If the radius of the small hoop is b and its mass is m and the radius of the large hoop =3b and its mass is 3m.
    My attempt: If x=0 is the present position of the hoops then their center of mass is x_cm = -b/2. I=I1+I2=3*m*(3b)^2 + m*b^2 = 28*m*b^2.
    Ug = Ktrans +Krot
    4*m*g*b/2 = .5*4*m*v^2 + .5*I*v^2/(-b/2)^2
    2*m*g*b = 2*m*v^2 + 2*28*b^2*v^2/b^2
    = 2*m*v^2 + 56*m*v^2. Therefore 29*v^2=g*b. Therefore v = (g*b/29)^.5;
    Is this the correct answer, if not where did I go wrong? Thanks.
  2. jcsd
  3. Oct 23, 2006 #2


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    I think you have a problem with your moment of inertia and relating the v you calculated to what was asked. If you are treating the two loops as one object and looking for the velocity of the CM, then you need I about the CM. Then you need to recognize that vCM is not the velocity of the center of the big hoop. I doubt that doing neither of these things leads to the correct answer.

    You could do separate calculations of energies of the two hoops connecting their velocities at the moment in question by the geometry and solve for v_big, or you can continue to pursue the CM approach you started.
  4. Oct 24, 2006 #3
    I'll try to pursue the cm first and see where it gets me.1st;
    Torque = I*(alpha), i.e., 4*m*g*b/2=I*(alpha) => 2*m*g=I*a/b^2, Then I = 2*m*g*b^2/a. So I need another equation with a in it. And such an equation is not obvious to me.
  5. Oct 24, 2006 #4


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    Your energy approach was OK. You can find the moment of inertia about the center of mass using the parallel axis theorem. You just need to know the distance to the CM from a point where you can calculate I. You know how to calculate I for each hoop at its center (that is what you did earlier) so you can find the I for each hoop about the CM of the combination of the two. The net I is just the sum of these two contributions.

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