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3 years ago

7

A teacup on a spinning teacup ride takes 1.5 s to complete a revolution around the center of the platform. What is the centripet

al acceleration of the teacup if it is 3.0 m from the center of the ride?

2 answers:

deff fn [24]3 years ago

6 0

Answer:

Centripetal acceleration, 52.6 m/s²

Explanation:

It is given that,

A teacup on a spinning teacup ride takes 1.5 s to complete a revolution around the center of the platform.

Distance covered by the particle in circular path is, $d=2\pi r$

We have to find centripetal acceleration of the teacup, $a_c=\dfrac{v^2}{r}$

Where, v is the velocity $v=\dfrac{2\pi r}{t}$

$a_c=\dfrac{4\pi^2 r}{t^2}$

$a_c=\dfrac{4\pi^2\times 3}{(1.5)^2}$

$a_c=52.6\ m/s^2$

Hence, centripetal acceleration of the teacup if it is 3.0 m from the center of the ride is 52.6 m/s²

arlik [135]3 years ago

5 0

The correct answer is 53 meters per second squared

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