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March 6, 2015

March 6, 2015

Posted by **darrin** on Sunday, March 3, 2013 at 4:56pm.

calculate the minimum speed needed to progress over a circular bump.

- physics -
**Damon**, Sunday, March 3, 2013 at 5:08pmat the top of the bump it stops rolling and stops going forward (teeters motionless)

Thus at the top the kinetic energy is zero and the potential energy is m g (r+h)

At the bottom the potential energy is m g r

so it gained potential energy of m g h

That is how much kinetic energy it lost

Ke = .5 m v^2 + .5 I w^2

I = m r^2

w = v/r

so in terms of v

Ke = .5 m v^2 + .5 m r^2 (v^2/r^2)

Ke = m v^2 total (half translational and half rotational)

so

m v^2 = m g h

v = sqrt (g h)

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