Question

A wheel's angular acceleration is constant. Initially its angular velocity is zero. During the first \(1.0-\mathrm{s}\) time interval, it rotates through an angle of \(90.0^{\circ} .\) (a) Through what angle does it rotate during the next 1.0 -s time interval? (b) Through what angle during the third 1.0 -s time interval?

Short answer

Answer: During the first 1.0-s time interval, the wheel rotates through an angle of \(\frac{\pi}{2}\, \mathrm{rad}\). During the second 1.0-s time interval, it rotates through an angle of \(\frac{3\pi}{2}\, \mathrm{rad}\). During the third 1.0-s time interval, it rotates through an angle of \(\frac{5\pi}{2}\, \mathrm{rad}\).

Step-by-step solution

(a) Calculate the angular acceleration during the first 1.0-s time interval

We have to find the angular acceleration \(\alpha\) during the first 1.0-s time interval. We are given that the angle rotated during this time interval is \(\theta = 90.0^\circ\). We will use the formula \(\theta = \omega_0 t + \frac{1}{2} \alpha t^2\), where \(\omega_0 = 0\), \(\theta = 90.0^\circ\), and \(t = 1.0 \, \mathrm{s}\). Converting degrees to radians, we get \(\theta = \frac{\pi}{2}\, \mathrm{rad}\). So, \(\frac{\pi}{2} = \frac{1}{2} \alpha (1)^2\). We can solve for the angular acceleration \(\alpha\): \(\alpha = \pi\, \mathrm{rad/s^2}\).

(b) Calculate the angle rotated during the second 1.0-s time interval

Now, we need to find the angle rotated during the second 1.0-s time interval. To do this, we will calculate the angle rotated during the first 2.0-s and subtract the angle rotated during the first 1.0-s. Using the formula \(\theta = \omega_0 t + \frac{1}{2} \alpha t^2\) with the calculated angular acceleration, \(\alpha = \pi\, \mathrm{rad/s^2}\), and time \(t = 2.0 \, \mathrm{s}\), we get: \(\theta_{2.0} = \frac{1}{2}\pi (2)^2 = 2\pi\, \mathrm{rad}\). Now, we subtract the angle rotated during the first 1.0-s (\(\theta_1 = \frac{\pi}{2}\, \mathrm{rad}\)) to find the angle rotated during the second 1.0-s time interval: \(\theta_{2} = \theta_{2.0} - \theta_{1} = 2\pi - \frac{\pi}{2}= \frac{3\pi}{2}\, \mathrm{rad}\).

(c) Calculate the angle rotated during the third 1.0-s time interval

Finally, we need to find the angle rotated during the third 1.0-s time interval. We will do this similarly to the second 1.0-s time interval by calculating the angle rotated during the first 3.0-s and subtracting the angle rotated during the first 2.0-s. Using the formula \(\theta = \omega_0 t + \frac{1}{2} \alpha t^2\) with the calculated angular acceleration, \(\alpha = \pi\, \mathrm{rad/s^2}\), and time \(t = 3.0 \, \mathrm{s}\), we get: \(\theta_{3.0} = \frac{1}{2}\pi (3)^2 = \frac{9\pi}{2}\, \mathrm{rad}\). Now, we subtract the angle rotated during the first 2.0-s (\(\theta_{2.0} = 2\pi\, \mathrm{rad}\)) to find the angle rotated during the third 1.0-s time interval: \(\theta_{3} = \theta_{3.0} - \theta_{2.0} = \frac{9\pi}{2} - 2\pi= \frac{5\pi}{2}\, \mathrm{rad}\). So, during the second 1.0-s time interval, the wheel rotates through an angle of \(\frac{3\pi}{2}\, \mathrm{rad}\), and during the third 1.0-s time interval, it rotates through an angle of \(\frac{5\pi}{2}\, \mathrm{rad}\).