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, (a) at what angular velocity (in rad/s) will the riders be subjected to a centripetal acceleration 1.5 times that due to gravity? 1.36 rad/s Correct (100.0%) Input 4 StatusYou have completed this input.Correct (100.0%) (b) How many revolutions per minute is this angular velocity equivalent to? 12.9 rpm Correct (100.0%) Input 5 StatusYou have completed this input.Correct (100.0%) (c) What is the magnitude of the centripetal force (in Newtons) on a 48.0 kg rider?

Answer 1

a)
`\omega=1.36rad/s`

b)
`\omega=12.99rpm`

c)
`F=705.6N`

Step-by-step explanation:

a) The angular velocity is related to the centripetal acceleration by the formula
`a_(cp)=\omega^2r`, which for our purposes we will write as:

`\omega=\sqrt{(a_(cp))/(r)}`

Since we want this acceleration to be 1.5 times that due to gravity, for our values we will have:

`\omega=\sqrt{(1.5g)/(r)}=\sqrt{((1.5)(9.8m/s^2))/((8m))}=1.36rad/s`

b) 1 rpm (revolution per minute) is equivalent to an angle of
`2\pi` radians in 60 seconds:

`1\ rpm=(2\pi rad)/(60s) =(\pi)/(30)rad/s`

Which means we can use the conversion factor:

`(1\ rpm)/((\pi)/(30)rad/s)=1`

So we have (multiplying by the conversion factor, which is 1, not affecting anything but transforming our units):

`\omega=1.36rad/s=1.36rad/s((1\ rpm)/((\pi)/(30)rad/s))=12.99rpm`

c) The centripetal force will be given by Newton's 2nd Law F=ma, so on the centripetal direction for our values we have:

`F=ma=(48kg)(1.5)(9.8m/s^2)=705.6N`