A tetherball apparatus consists of a ball attached to a 2.10 m rope that is atta

Question

A tetherball apparatus consists of a ball attached to a 2.10 m rope that is attached in turn to the top of a 2.70 m tall pole. The ball is struck such that it moves horizontally around the pole. Immediately after the ball is struck, it moves at 5.57 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?

I got a height of 1.65 m off the ground.

If you start over and hit the ball harder, such that it moves with speed 16.9 m/s at a height of 2.55 m, what would the centripetal acceleration of the ball have to be (as a multiple of g)?

Explanation / Answer

dy = 2.7 - 2.55

so r = sqrt(2.1^2 - (2.7-2.55)^2)

a = v^2/r = 16.9^2/ sqrt(2.1^2 - (2.7-2.55)^2)

=13.9 g

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.