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Linear speed up a ramp

  1. Oct 23, 2006 #1
    Linear speed up a "ramp"

    After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of v = 3.02 m/s (Figure 10-24). To reach the rack, the ball rolls up a ramp that rises through a vertical distance of h = 0.47 m. What is the linear speed of the ball when it reaches the top of the ramp?

    I tried using linear equations, specifically v^2=vo^2 + 2gy. This homework problem was in the same set of problems concerning rotational kinematics, so I'm not sure if that is supposed to play a part in it or not.
     
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  3. Oct 23, 2006 #2

    berkeman

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    Staff: Mentor

    You need to take into account the rotational energy of the ball. What is the equation relating the moment of inertia of an object to its rotational energy? What is the equation for the moment of inertia of a uniform sphere? You can use energy to solve most of this problem. The ball has some initial rotational and translational kinetic energy as it rolls along the flat plane, and then it gains PE and loses both rotational and linear KE as it goes up the ramp.
     
  4. Oct 23, 2006 #3
    The moment of inertia for a sphere is I=2/5 mr^2 and rotational kinetic energy is KE=1/2 mv^2+1/2 Iw^2. By substituting I into the KE equation, I got KE=1/2 mv^2 + 1/2(2/5 mv^2) and by symplifying, I got KE=1/2 mv^2 + 1/5 mv^2. Using conservation of energy, I did KE(initial)+PE (initial)= KE (final) + PE (final). In specific: 1/2 vo^2 + 1/5 vo^2 = 1/2 v^2 + 1/5 v^2 +gh and then I solved for v.
    ...am I somewhat on the right track??
     
  5. Oct 23, 2006 #4

    berkeman

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    Looks good so far! It's easy to overlook the rotational energy term in problems like this, so keep an eye out for things with changing rotational energy, and keep track of that term.
     
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