Respuesta :
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, [tex]d=2\pi r[/tex]
We have to find centripetal acceleration of the teacup, [tex]a_c=\dfrac{v^2}{r}[/tex]
Where, v is the velocity [tex]v=\dfrac{2\pi r}{t}[/tex]
[tex]a_c=\dfrac{4\pi^2 r}{t^2}[/tex]
[tex]a_c=\dfrac{4\pi^2\times 3}{(1.5)^2}[/tex]
[tex]a_c=52.6\ m/s^2[/tex]
Hence, centripetal acceleration of the teacup if it is 3.0 m from the center of the ride is 52.6 m/s²
