Question
A child swings a tennis ball attached to a 0.662-m string in a horizontal circle above his head at a rate of 4.50 rev/s.
(a) What is the centripetal acceleration of the tennis ball?
m/s2
(a) What is the centripetal acceleration of the tennis ball?
m/s2
Answers
omega^2 R
omega = 4.5 revs/s * 2 pi radians/rev
= 9 pi rad/s
omega^2 = 81 pi^2 /s^2
so
a = 81 pi^2 * .662
= 529 m/s^2
omega = 4.5 revs/s * 2 pi radians/rev
= 9 pi rad/s
omega^2 = 81 pi^2 /s^2
so
a = 81 pi^2 * .662
= 529 m/s^2
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