patrons on the midway ride called the rotor stand with their backs against the wall of a revolving cylinder…

patrons on the midway ride called the rotor stand with their backs against the wall of a revolving cylinder, while the floor drops away from beneath them. to keep from falling, they require a centripetal acceleration of about 25 m/s2. if the rotor has a diameter of 5.0 m, with what minimum frequency does it revolve. input your answer in terms of hz or 1/s but do not include the unit in your answer. round to one decimal place. answer:

patrons on the midway ride called the rotor stand with their backs against the wall of a revolving cylinder, while the floor drops away from beneath them. to keep from falling, they require a centripetal acceleration of about 25 m/s2. if the rotor has a diameter of 5.0 m, with what minimum frequency does it revolve. input your answer in terms of hz or 1/s but do not include the unit in your answer. round to one decimal place. answer:

Answer

Explanation:

Step1: Find the radius

The diameter $d = 5.0$ m, so the radius $r=\frac{d}{2}=\frac{5.0}{2}= 2.5$ m.

Step2: Use centripetal - acceleration formula

The centripetal - acceleration formula is $a = r\omega^{2}$, where $a = 25$ m/s² and $r = 2.5$ m. We can solve for $\omega$: $\omega=\sqrt{\frac{a}{r}}=\sqrt{\frac{25}{2.5}}=\sqrt{10}\text{ rad/s}$.

Step3: Convert angular frequency to linear frequency

The relationship between angular frequency $\omega$ and linear frequency $f$ is $\omega = 2\pi f$. So $f=\frac{\omega}{2\pi}$. Substitute $\omega=\sqrt{10}\text{ rad/s}$ into the formula: $f=\frac{\sqrt{10}}{2\pi}$. Calculate $f=\frac{\sqrt{10}}{2\pi}\approx\frac{3.162}{2\times3.142}\approx0.5$.

Answer:

0.5