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problem 2: a cd spins to a stop. its initial frequency is 200 revolutio…

Question

problem 2:
a cd spins to a stop. its initial frequency is 200 revolutions per minute.
given an angular deceleration of -15 rad/s², find the time for the cd to
stop spinning.

Explanation:

Step1: Convert initial frequency to angular velocity

The formula to convert frequency \(f\) (in revolutions per minute) to angular velocity \(\omega_0\) (in rad/s) is \(\omega_0 = 2\pi f/60\).
Given \(f = 200\) revolutions per minute, then \(\omega_0=\frac{2\pi\times200}{60}=\frac{20\pi}{3}\text{ rad/s}\approx 20.94\text{ rad/s}\).

Step2: Use the angular - motion formula

The angular - motion formula is \(\omega=\omega_0+\alpha t\). When the CD stops, \(\omega = 0\). We know \(\omega_0\) (from Step 1) and \(\alpha=- 15\text{ rad/s}^2\).
Substitute into the formula \(0=\omega_0+\alpha t\), then \(t =-\frac{\omega_0}{\alpha}\).
Substitute \(\omega_0=\frac{20\pi}{3}\text{ rad/s}\) and \(\alpha=-15\text{ rad/s}^2\) into \(t =-\frac{\omega_0}{\alpha}\), we get \(t=\frac{\frac{20\pi}{3}}{15}=\frac{20\pi}{45}=\frac{4\pi}{9}\approx1.40\text{ s}\).

Answer:

\(t\approx1.40\text{ s}\)