A tetherball apparatus consists of a ball attached to a 1.90 m rope that is atta

Question

A tetherball apparatus consists of a ball attached to a 1.90 m rope that is attached in turn to the top of a 3.10 m tall pole. The ball is struck such that it moves horizontally around the pole. Immediately after the ball is struck, it moves at 5.30 m/s, and experiences a centripetal acceleration of 1.74g (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 15.3 m/s at a height of 2.95 m, what would the centripetal acceleration of the ball have to be (as a multiple of g)?

Explanation / Answer

acceleration = a = v^2/r

a = 5.3^2 / 1.9

a = 14.78

centripetal acceleration = a*sintheta = 14.78 (h/1.9)

hence

14.78 (h/1.9) = 1.74g

hence

h = 2.192 m

a = v^2/r = 15.3^2/1.9 = 123.21

centripetal acceleration = 123.21(2.95/1.9) = 191.3 = 19.52g

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