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March 4, 2015

March 4, 2015

Posted by **jeff** on Sunday, December 21, 2008 at 12:09pm.

How high above the table must the rolling ball start to just make the loop?

How far does the ball travel along the loose length of track before the track starts to tip?

- physics -
**drwls**, Sunday, December 21, 2008 at 1:01pmThe mass of the ball will not matter in the first part. The ball must be released high enough to achieve a velocity V such that V^2/R = g at the top of the loop. If H is the initial height above the table,

M g H - M g (2R)

= (1/2) M V^2 + (1/2)*(2/5)M V^2

= (7/10)M V^2

The second kinetic energy term is due to rotation of the spherical ball. M's cancel. Solve for H

For the last question, the velocity does not matter, but M does. The track starts to tip when the moment due to the ball about the edge of the table is equal and opposite to the moment due to the weight of the track, acting at the CG.

- physics -
**Anonymous**, Sunday, January 15, 2012 at 12:30amTwo balls are conneted by a string that stretches over a massless, frictionless pulley. Ball 1 has a mass of 0.81 kg and is held 0.5 m above the ground. Ball 2 has a mass of 6.3 kg and is held 0.28 m above the ground. When the balls are released, ball 2 falls to the ground, looses 29 \% of its kinetic energy and rebounds to some maximum rebound height. When the balls are released, ball 1 travels to some maximum height before starting to fall. Assume that ball 1 reaches its maximum height during ball 2's rebound so that the string doesn't pull.

Calculate the maximum height of ball 1 from and ground and the rebound height of ball 2.

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