A turntable must spin at 33.3 rpm \((3.49 \mathrm{rad} / \mathrm{s})\) to play an old-fashioned vinyl record. How much torque must the motor deliver if the turntable is to reach its final angular speed in 2.0 revolutions, starting from rest? The turntable is a uniform disk of diameter \(30.5 \mathrm{cm}\) and mass \(0.22 \mathrm{kg}\).

To find the moment of inertia, we need to use the following formula for a uniform disk: \(I = \frac{1}{2}mr^2\) where \(I\) is the moment of inertia, \(m\) is the mass of the disk, and \(r\) is the radius. In this case, we are given the mass of the turntable as \(0.22 kg\) and the diameter as \(30.5 cm\). First, we need to find the radius by dividing the diameter by 2: \(r = \frac{30.5}{2} = 15.25 cm = 0.1525 m\) Now, plug the mass and radius into the formula: \(I = \frac{1}{2}(0.22)(0.1525)^2 = 0.00257 kg * m^2\)

The turntable starts from rest, so the initial angular speed is 0. We need to find the angular acceleration such that it reaches its final angular speed, \(3.49 rad/s\), in just 2.0 revolutions. First, let's convert the number of revolutions to radians. There are \(2\pi\) radians in 1 revolution, so: \(2.0 \text{ revolutions} \times \frac{2\pi \mathrm{rad}}{1 \text{ revolution}} = 4\pi \mathrm{rad}\) Now, we can use the final angular displacement formula: \(\omega_f^2 = \omega_i^2 + 2\alpha\theta\) where \(\omega_f\) is the final angular speed, \(\omega_i\) is the initial angular speed, \(\alpha\) is the angular acceleration, and \(\theta\) is the angular displacement in radians. Since the turntable starts from rest, \(\omega_i = 0\). Plugging the known values into the equation: \((3.49)^2 = 0 + 2\alpha(4\pi) \Longrightarrow \alpha = \frac{(3.49)^2}{8\pi} = 0.612 \mathrm{rad/s^2}\)

Now that we have the moment of inertia and the angular acceleration, we can calculate the torque using the following formula: \(\tau = I\alpha\) Plugging in the values for \(I\) and \(\alpha\): \(\tau = (0.00257)(0.612) = 0.00157 \mathrm{kg \cdot m^2/s^2}\) The torque required for the turntable to reach its final angular speed in 2.0 revolutions is approximately \(0.00157 \mathrm{kg \cdot m^2/s^2}\).