Question

A teacup on a spinning teacup ride takes 1.5 s to complete a revolution around the center of the platform. What is the centripetal acceleration of the teacup if it is 3.0 m from the center of the ride?

Answer 1

The correct answer is 53 meters per second squared 

Answer 2

Centripetal acceleration, 52.6 m/s²

Step-by-step 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=(v^2)/(r)`

Where, v is the velocity
`v=(2\pi r)/(t)`

`a_c=(4\pi^2 r)/(t^2)`

`a_c=(4\pi^2* 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