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Posted by on Saturday, April 12, 2008 at 4:52pm.

The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center. When the skater's hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thin-walled hollow cylinder. His hands and arms have a combined mass of 8.50kg. When outstretched, they span 1.80m; when wrapped, they form a cylinder of radius 25.0cm. The moment of inertia about the axis of rotation of the remainder of his body is constant and equal to 0.450 kg·m^2. If the skater's original angular speed is 0.450 rev/s, what is his final angular speed?

  • physics - , Saturday, April 12, 2008 at 6:59pm

    The total moment of inertia with arms outstretched is:
    I1 = Ibody + Iarms = 0.450 + (1/12) m L^2 = 0.450 + (1/12)*8.5*(1.8)^2 = 2.75 kg m^2

    The total moment of inertia with arms rapped aound the body is:
    I2 = Ibody + Iarms = 0.450 + m R^2 =
    0.45 + 0.53 = 0.98 kg m^2

    Use the conservation of angular momentum equation
    I1*w1 = I2*w2 to get the final anglualr speed w2. It should increrase by about a factor of 3.

  • physics - , Sunday, March 22, 2009 at 8:31pm

    drwls thanks so much. your solution is absolutely right haha =) i got a similar qns to that of Erin too. and managed to solve it thanks to u.

  • physics - , Wednesday, December 4, 2013 at 2:01am


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