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Homework Help: Physics help- unsure where I went wrong

  1. Apr 14, 2009 #1
    1. A thin, uniform, metal bar, 2.5 m long and weighing 90 N, is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.3 m below the ceiling by a small 4-kg ball, initially traveling horizontally at 14 m/s. The ball rebounds in the opposite direction with a speed of 7 m/s.

    Part A. Find the angular speed of the bar just after the collision.

    I tried to solve using conservation of angular momentum (it must be a calculation error but I keep getting the same answer):
    m1v_0d = -m1vd +(1/3)m2L^2(omega)
    (4.0kg)(14m/s)(1.3m) = -(3kg)(7m/s)(1.3m)+(1/3)(90N/9.82m/s)(2.5m)^2(omega)
    72.8 = 57.398 + (-27.3)
    1001.1 = 57.398
    1.7439rad/s = \omega

    2. A Ball Rolling Uphill. A bowling ball rolls without slipping up a ramp that slopes upward at an angle beta to the horizontal. Treat the ball as a uniform, solid sphere, ignoring the finger holes.
    Part A. What is the acceleration of the center of mass of the ball?

    I tried drawing it out, f = mew N, N = wsin(beta)
    - (WsinF+ muWcos(beta) = macm
    I entered acm = -g(sinF+mucos (beta)
    It was incorrect, so I thought I over-did it, and tried just -g(sin(beta)), but that was also incorrect, I'm not sure where to go from here.

    Part B. What minimum coefficient of static friction is needed to prevent slipping?
    I tried mu = tan (beta)
  2. jcsd
  3. Apr 14, 2009 #2


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    Homework Helper


    You have

    |r X P| = r*m*vi = r*m*(-vf) + I*ω

    So r*m*(vi + vf) = 1.3*4*(14 + 7) = 1.3*4*21 = 109.2 = 1/3*90/9.8*L2

    So then ...

    ω = 109.2*9.8*3/(90*2.5*2.5) = ... ?
  4. Apr 14, 2009 #3
    thanks!- I've got a calculator now (much easier! Mine stopped working last week so I've been borrowing from a friend), 5.70752.

    Any tips on the next question?
  5. Apr 14, 2009 #4


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    For 2) consider the torque about the center of the solid sphere. That might yield "a" angular acceleration which you can then relate to linear acceleration along the incline.
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