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Torque/angular momentum/etc questions

  1. Nov 22, 2008 #1
    I'm studying for a test (YIKES!) by doing practice mc questions, which unfortunately do not have the answers to or don't understand the answer. Can someone(s) check my answers and explain what I did wrong?



    A rod of negligible mass is pivoted at a point that is off center, so that length l1 is different from length l2. The figures above show two cases in which masses are suspended from the ends of the rod. In each case the unknown mass m is balanced by a known mass, M1 or M2, so that the rod remains horizontal. What is the value of m in terms of the known masses?


    The answer is E, but I have no idea where that came from. So far I have

    ml2= M1l2
    M2l1 = ml2

    Where do I go from there?

    The rest of them I don't have the answer key.

    For 2 and 3


    A sphere of mass M, radius r, and rotational inertia I is released from rest at the top of an inclined plane of height h as shown above.

    2. If the plane is frictionless, what is the speed vcm of the center of mass of the sphere at the bottom of the incline?


    A because it the plane is friction and it slips so all the energy is translational.

    3. If the plane has friction so that the sphere rolls without slipping, what is the speed vcm of the center of mass at the bottom of the incline?

    Same set of answers as #2. This one I don't understand.

    I think I used use U = Ktrans + Krot but still can't rearrange it to get one of the answer choices...

    4. A wheel of 0.5 m radius rolls without slipping on a horizontal surface. The axle of the wheel advances at constant velocity, moving a distance of 20 m in 5 s. The angular speed of the wheel about its point of contact on the surface is

    (A) 2 radians s-1
    (B) 4 radians s-1
    (C) 8 radians s-1
    (D) l6 radians s-1
    (E) 32 radians s-1

    I solved for v, used the equation v = ωr and got c as the answer.


    5. A long board is free to slide on a sheet of frictionless ice. As shown in the top view above, a skater skates to the board and hops onto one end, causing the board to slide and rotate. In this situation, which of the following occurs?

    A) Linear momentum is converted to angular momentum.
    (B) Kinetic energy is converted to angular momentum.
    (C) Rotational kinetic energy is conserved.
    (D) Translational kinetic energy is conserved.
    (E) Linear momentum and angular momentum are both conserved.

    I'ts A. I can't explain well but the other answers look wrong.


    6. Two blocks are joined by a light string that passes over the pulley shown above, which has radius R and moment of inertia I about its center. T1 and T2 are the tensions in the string on either side of the pulley and a is the angular acceleration of the pulley. Which of the following equations best describes the pulley’s rotational motion during the time the blocks accelerate?

    (A) m2 gR = Iα
    (B) (T1+T2) R = Iα
    (C) T2 R = Iα
    (D) (T2 - T1) R = Iα
    (E) (m2 - m1) gR = Iα

    It's D.

    Please help.
  2. jcsd
  3. Nov 22, 2008 #2
    Let's do one at a time.

    Clearly, you need to eliminate l1 and l2 from these equations. What's your best attempt at that?
  4. Nov 22, 2008 #3
    Problem 3, what do you have?
  5. Nov 22, 2008 #4
    For 1 I isolated l1, set it equal to itself, solved for m, and got E. This one's done. :)

    For 3.

    Mgh = (1/2)Mv2 + (1/2)Iv2/r2

    I tried to isolate v but couldn't match one of the answer choices.
  6. Nov 22, 2008 #5
    This is correct. Find an expression for v, write it down, and eliminate the choices. You will get it.
  7. Nov 22, 2008 #6
    5. Can you explain why the others are wrong?
  8. Nov 22, 2008 #7
    4. I'm confused by what "angular velocity of the wheel about the point of contact" means. Maybe I'll let someone else answer that. I would think you are right.

    6. What's your question?
  9. Nov 22, 2008 #8
    I guess 3 is E. They multiplied by the terms by r2... didn't expect that.

    4. They like to word things in a confusing way. I'm assuming they mean the angular velocity at that point.

    5. The others are wrong because kinetic energy is not conserved in inelastic collisions.

    6. I used the equation Στ = Iα

    Στ = RT2 - RT1

    That's where I got the answer from. Is it right?
  10. Nov 22, 2008 #9
    5. Yes, kinetic energy is not conserved. Can you explain why that makes each of B, C, D and E false?

    6. Correct.
  11. Nov 22, 2008 #10
    2) use conservation of energy, the energy at the beginning is the same as the energy at the end, and since there is no friction the ball does not roll, so it is the same as if a block were sliding down the plane.

    3) Check out the beginning of this lecture:
    http://ocw.mit.edu/OcwWeb/Physics/8-01Physics-IFall1999/VideoLectures/detail/embed24.htm [Broken]
    Last edited by a moderator: May 3, 2017
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