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Rotational & Translational Energy

  1. Nov 8, 2007 #1
    A solid, uniform, spherical boulder starts from rest and rolls down a 50.0-m high hill. The to half of the hill is rough enough to cause the boulder to roll without slipping, but the lower half is covered with ice and there is no friction. Wat is the translational speed of the boulder when it reaches the bottom of the hill?

    Attempt at solving: Energy is conserved, so the potential energy (mgh) at the top is equal to the combination of the kinetic and translational energy as it rolls down the rough part. Solving for that, I get v = sqrt(10/7*gh), or v = 26.46 m/s. But that's only part of the problem, and I'm not sure what to do next. It would seem to me the energy on the icy part is only translational, not rotational, and maybe the eqn would be 26.46 m/s = 1/2*m*v2^2, but mass is not given. Any help would be very much appreciated.
     
  2. jcsd
  3. Nov 8, 2007 #2

    Doc Al

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    Staff: Mentor

    In each part, the loss in PE equals the increase in KE. In the first have, the KE is shared between translational and rotational KE. But in the second half, only the translational KE increases. (You don't need the mass.)
     
  4. Nov 11, 2008 #3
    How exactly do you solve this, though? I keep getting a velocity value that is off by quite a bit.

    Do you set 0.5mv^2 = the KE at the halfway point as well as the PE at the halfway point?

    It does not work.
     
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