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A couple problems with angular momentum

  1. Nov 10, 2003 #1
    I know a couple of these I already asked, but I'm really close and missing something. Could someone help??

    1) A ring of mass 2.53 kg, inner radius 6.00 cm, and outer radius 8.00 cm is rolling (without slipping) up an incline plane which makes an angle of = 36.4°. At the moment the ring is at position x = 2.00 m up the plane, its speed is 2.75 m/s. The ring continues up the plane for some additional distance, and then rolls back down. It does not roll off the top end. How far up the plane does it go?

    I = (1/2)*M*(R1^2+R2^2) = .01265 kg m^2.
    v = rw
    2.75 m/s = .08mw
    w=34.375 rad/sec

    w = omega.

    (1/2)mv^2+(1/2)Iw^2 = mgh (I'm setting the initial mgh to zero.)
    (1/2)(2.53kg)(2.75m/s)^2 + (1/2)(.01265 kg m^2)(34.38)^2 = (2.53kg)(9.8m/s^2)(sin 36.4)x
    x = 1.158 m

    initial distance up at the x = 2 mark on the plane is 1.19m.

    1.158 + 1.19 = 2.35 m

    What's wrong??

    2) A uniform solid disk of mass 2.98 kg and radius 0.200 m rotates about a fixed axis perpendicular to its face. Omega = 5.95 rad/sec.
    What is the angular momentum when the axis of rotation passes through a point midway between the center and the rim?

    I = I(i)+mr^2
    I = .355kgm^2+(2.98kg)(.1^2)
    I = 2.30 kg m^2 / sec

    What's wrong???

    3) A solid sphere of mass m and radius r rolls without slipping along the track. The sphere starts from rest with the lowest point of the sphere at height h above the bottom of the loop of radius R, which is much larger than r.

    What are the force components on the sphere at the point P if h = 3R?

    I got Fx being -20/7mg, and I thought Fy would be -mg but it isn't.

    THanks.
     
  2. jcsd
  3. Nov 10, 2003 #2

    jamesrc

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    Science Advisor
    Gold Member

    1. At first glance, your solution looks good.

    Is it possible that when they say 2 m up the plane, they mean along the plane (making your final answer 2 + 1.158)?

    2. Be careful here. I think you meant to write:

    Io = .5*M*R^2 = .0596 kgm^2

    I = .5*M*R^2 + Mh^2 = .0596 + 2.98*(.1)^2

    andular momentum L = Iω = I*5.95rad/s (=.53 kg m^2/s)

    3. Where is point P? Somewhere on the loop? If you get a chance, could you show where your answer came from too?
     
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