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quantumlolz
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Homework Statement
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]
Homework 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?
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.