Question

A flying saucer shaped fairground ride is rotating in a horizontal plane. If the rider’s circular path has a radius of 8m (r=8m) then at how many revolutions per minute should the ride spin in order for the rider to feel a centripetal acceleration of about 1.5 times Earth’s gravitational acceleration?

Answer 1

Before heading towards the solution one must know that revolution per second is known as the frequency of that rotating object. Now, in order to solve this question we will approach it with the formula for centripetal acceleration and then relate acceleration with frequency of the fairground ride and find at how many revolutions per minute the ride should spin in order for the rider to feel a centripetal acceleration of about $1.5$ times the Earth’s gravitational acceleration.
$F=ma=\dfrac{m{{v}^{2}}}{r}$
Where $F$ is the centripetal force,
${{a}_{c}}$ is the centripetal acceleration,
$m$ is the mass of the object,
$v$ is the tangential velocity of the object,
And $r$ is the radius (distance of the object from the center).
$v=r\varpi$
Where, $\varpi$ is the angular velocity of the object about the center.
$\varpi =2\pi f$
Where, $f$ is the frequency in the number of cycles per second.

Complete step by step answer:
For circular motion the magnitude of the centripetal force $F$ of an object with mass $m$ moving along a path with radius $r$ with tangential velocity $v$ is given by
$F=ma=\dfrac{m{{v}^{2}}}{r}...(1)$
Where ${{a}_{c}}$ is the centripetal acceleration.
The tangential velocity $v$ is related to the angular velocity $\varpi$ of the object about the center of the circular path by the expression
$v=r\varpi$
And the angular velocity $\varpi$ of the object is also related to frequency $f$ in number of cycles per second of the object as
$\varpi =2\pi f$
So now,
${{a}_{c}}=\dfrac{{{v}^{2}}}{r}=r{{\varpi }^{2}}...\left( 2 \right)$ from $\left( 1 \right)$
$\Rightarrow {{a}_{c}}=r{{\left( 2\pi f \right)}^{2}}$
$\Rightarrow 4{{\pi }^{2}}{{f}^{2}}r={{a}_{c}}...\left( 3 \right)$ from $\left( 2 \right)$
Here, we are asked to find frequency per minute for the fairground ride with centripetal acceleration equal to $1.5g$ .
So by solving equation $\left( 3 \right)$ for $f$ per minute we will also multiply it by $60$ and inserting the known values we get
$f=\sqrt{\dfrac{{{a}_{c}}}{4{{\pi }^{2}}r}}\times 60$
$f=\sqrt{\dfrac{1.5\times 9.81}{4{{\pi }^{2}}\times 8}}\times 60\approx 13$ .

Note:
One must not get confused between linear motion and circular motion. This problem was of circular motion so it had angular velocity and tangential velocity both whereas in linear motion problems there is no angular velocity and only linear velocity. Here another thing to keep in mind was to multiply the frequency with 60 so as to convert it from revolutions per second to revolutions per minute.