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

Answer:

a)$\omega=1.36rad/s$

b)$\omega=12.99rpm$

c)$F=705.6N$

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:

$\omega=\sqrt{\frac{a_{cp}}{r}}$

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

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

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

$1\ rpm=\frac{2\pi rad}{60s} =\frac{\pi}{30}rad/s$

Which means we can use the conversion factor:

$\frac{1\ rpm}{\frac{\pi}{30}rad/s}=1$

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

$\omega=1.36rad/s=1.36rad/s(\frac{1\ rpm}{\frac{\pi}{30}rad/s})=12.99rpm$

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

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