Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Conservation of energy, Rolling without slipping.

  1. Apr 11, 2010 #1
    1. The problem statement, all variables and given/known data

    A bowling ball is rolling without slipping up an inclined plane. As it passes a point O it has a speed of 2.00 m/s up the plane. It reaches a vertical height h above O before momentarily stopping and rolling back down. Determine the value of h. The moment of inertia of a solid sphere of mass M and radius R is [tex]I = \frac{2MR^2}{5}[/tex]

    2. Relevant equations

    Translational kinetic energy [tex] \frac{mv^2}{2}[/tex]
    Rotational kinetic energy [tex] \frac {I\omega^2}{2} [/tex]
    Rolling without slipping [tex] v = r\omega [/tex]
    Gravitational potential energy [tex] U = mgy [/tex]
    Constant acceleration formulae maybe?

    3. The attempt at a solution

    Equating Total (rotational and translational) kinetic energy and potential energy [tex] \frac{Mv^2}{2} + \frac {I\omega^2}{2} = Mgy [/tex] as all the energy has been converted from kinetic to potential when the ball stops (i.e. at the height we're trying to find)

    Inserting the expression for moment of inertia the rotational kinetic energy is [tex] \frac{MR^2\omega^2}{5} [/tex] but for rolling without slipping this is just [tex] \frac{Mv^2}{5} [/tex] so total kinetic energy becomes [tex] T = \frac{Mv^2}{2} + \frac{Mv^2}{5} = \frac {7Mv^2}{10} [/tex]

    So [tex] \frac{7Mv^2}{10} = Mgy [/tex] and we can cancel the Ms to get [tex] \frac {7v^2}{10} = gy [/tex]

    At some value of y (lets call it d), v = 2.00 m/s so plugging in the numbers [tex] d = \frac{7v^2}{10g} [/tex] = 0.29m

    Then I'm not sure what to do next. I was thinking we've got an initial velocity (2 m/s), a final velocity (0 m/s), an acceleration (g), and we want the vertical displacement, h, from O to give a final velocity of 0 m/s so we can then apply constant acceleration equation [tex] v^2 = u^2 + 2gh [/tex], with v = 2, u = 0 and solving this gives [tex] h = \frac{v^2}{2g} [/tex] but this comes to h = 0.20m, the quoted answer is h = 0.29 m.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Apr 11, 2010 #2


    User Avatar
    Homework Helper

    You got the correct result, h=0.29 m, only you did not notice.

    You can take the potential equal to zero at any point, here a good choice is at point O.

    At point O the ball has vo=2m/s velocity and it rolls upward. You calculated the KE with this velocity, correctly. This KE transforms to potential energy h height above O, where the potential energy with respect to O, is mgh. As you wrote,

    d = \frac{7v^2}{10g} =0.29 m

    But d is the height h which was the question. You got it!

  4. Apr 12, 2010 #3
    Ah cool, I did spot that I got the right number when I was working it out, but wasnt sure why - turns out I'd completely forgotten that I can choose where the potential is zero. Thanks :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook