A tetherball apparatus consists of a ball attached to a 2.30 m rope that is atta

Question

A tetherball apparatus consists of a ball attached to a 2.30 m rope that is attached in turn to the top of a 2.90 m tall pole. The ball is struck such that it moves horizontally around the pole. Immediately after the ball is struck, it moves at 2.56 m/s, and experiences a centripetal acceleration of 0.580g (where g is the acceleration due to gravity of a falling object near Earth's surface). If the resulting path of the ball is approximately a circle, what is the height h of the ball above the ground?

If you start over and hit the ball harder, such that it moves with speed 18.6 m/s at a height of 2.75 m, what would the centripetal acceleration of the ball have to be (as a multiple of g)?

ps. First answer is in m and second part is in g.

Explanation / Answer

a)

Given centripetal acceleration.

v^2 / r = g

2.56^2 / r = 0.580 * 9.8

r = 1.153 m

Height of the ball above the ground

= 2.90 - sqrt(2.30^2 - 1.153^2)

= 0.90 m

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