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

Answer:

The number of revolutions is 10.68 rev/min.

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

$a_{c}=\dfrac{v^2}{r}$

$v^2=a_{c}\times r$

Put the value into the formula

$v=\sqrt{9.8\times8}$

$v=8.85\ m/s$

We need to calculate the value of $\omega$

Using formula of velocity

$v=r\omega$

$\omega=\dfrac{v}{r}$

Put the value into the formula

$\omega=\dfrac{8.85}{8}$

$\omega=1.12\rad/s$

We need to calculate the number of revolutions

Using formula of angular frequency

$\omega=\dfrac{2\pi}{T}$

$\omega=2\pi N$

$N=\dfrac{\omega}{2\pi}$

Put the value into the formula

$N=\dfrac{1.12}{2\pi}$

$N=0.178\ rev/s$

Using conversion rev/s to rev/min

$N=0.178\times 60$

$N=10.68\ rev/min$

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