A merry-go-round completes one revolution in 40 s. What is the centripetal acceleration of a rider located at a distance 2 m from the center of the merry go-round? A) m/s² B) $\pi$ m/s² C) $\pi$²/200 m/s² D) $\pi$²/100 m/s² E) $\pi$² m/s²

Question

Quizplus · Accepted Answer

The formula for centripetal acceleration is a = v^2/r, where v is the tangential velocity and r is the distance from the center. We can determine the tangential velocity using the formula v = 2πr/T, where T is the period (time for one revolution) and π is a constant value of approximately 3.14. Plugging in the given values, we get:

v = 2π(2)/40 = 0.314 m/s

Now we can plug in the values for v and r into the formula for centripetal acceleration:

a = (0.314)^2/2 = 0.049 m/s^2

Simplifying, we get:

a = 1/200 m/s^2

Therefore, the best answer is C, which gives the centripetal acceleration as 1/200 m/s^2.

A Merry-Go-Round Completes One Revolution in 40 S π\piπ M/s2 C) π\piπ

A Merry-Go-Round Completes One Revolution in 40 S $\pi$ M/s²
C) $\pi$