Question

A flywheel of radius \(31.59 \mathrm{~cm}\) is rotating at a constant frequency. The centripetal acceleration at a point on the edge of the flywheel is \(8.629 \cdot 10^{4} \mathrm{~m} / \mathrm{s}^{2}\). What is the frequency of rotation of the flywheel (in rpm)?

Short answer

Answer: The frequency of rotation of the flywheel is approximately 4987.2 rpm.

Step-by-step solution

Write down the formula for centripetal acceleration

The formula for centripetal acceleration (a_c) at the edge of a rotating wheel is given by: $$ a_c = r \omega^2 $$ where \(r\) is the radius of the wheel and \(\omega\) is the angular velocity (measured in radians per second).

Convert given radius to meters

The radius \(r\) is given in centimeters, but the centripetal acceleration is given in \(\frac{m}{s^2}\). So we need to convert the radius into meters: $$ r = 31.59~cm \cdot \frac{1~m}{100~cm} = 0.3159~m $$

Substitute the values in the centripetal acceleration formula and solve for angular velocity

We have the centripetal acceleration, \(a_c = 8.629 \cdot 10^4 \frac{m}{s^2}\), and the radius, \(r = 0.3159~m\). Substituting these values into the formula, we get: $$ 8.629 \cdot 10^4 = 0.3159 \omega^2 $$ Now, we can solve for \(\omega\): $$ \omega^2 = \frac{8.629 \cdot 10^4}{0.3159} $$ $$ \omega = \sqrt{\frac{8.629 \cdot 10^4}{0.3159}} \approx 522.33~rad/s $$

Find the frequency (in Hz) from angular velocity

We can find the frequency (in Hz) from the angular velocity using the following relationship: $$ \omega = 2\pi f $$ where \(f\) is the frequency in Hz. Solving for \(f\): $$ f = \frac{\omega}{2\pi} $$ $$ f = \frac{522.33}{2\pi} \approx 83.12~Hz $$

Convert the frequency to rotations per minute (rpm)

To convert the frequency from Hz to rotations per minute (rpm), we use the following relationship: $$ 1~Hz = 60~rpm $$ So, the frequency of the flywheel in rotations per minute is: $$ f_{rpm} = 83.12 \cdot 60 \approx 4987.2~rpm $$ The frequency of rotation of the flywheel is approximately \(4987.2~rpm\).