Question

A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 8.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose mag

Answer 1

The number of revolutions is 10.68 rev/min.

Step-by-step explanation:

Given that,

Radius = 8 m

Suppose, centripetal acceleration equal to the gravity

`a_(c)=g=9.8`

We need to calculate the velocity

Using formula of centripetal acceleration

`a_(c)=(v^2)/(r)`

`v^2=a_(c)* r`

Put the value into the formula

`v=√(9.8*8)`

`v=8.85\ m/s`

We need to calculate the value of
`\omega`

Using formula of velocity

`v=r\omega`

`\omega=(v)/(r)`

Put the value into the formula

`\omega=(8.85)/(8)`

`\omega=1.12\rad/s`

We need to calculate the number of revolutions

Using formula of angular frequency

`\omega=(2\pi)/(T)`

`\omega=2\pi N`

`N=(\omega)/(2\pi)`

Put the value into the formula

`N=(1.12)/(2\pi)`

`N=0.178\ rev/s`

Using conversion rev/s to rev/min

`N=0.178* 60`

`N=10.68\ rev/min`

Hence, The number of revolutions is 10.68 rev/min.