Question

A turntable must spin at 33.3 rev/min (3.49 rad/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 2.60 revolutions, starting from rest? The turntable is a uniform disk of diameter 30.5 cm and mass 0.250 kg. N·m

Answer 1

Given data:

* The initial angular velocity of the turntable is,

$\omega_1=0\text{ rad/s}$

* The final angular velocity of the turntable is,

$\omega_2=3.49\text{ rad/s}$

* The mass of the turntable is m = 0.25 kg.

* The diameter of the turntable is D = 30.5 cm.

* The number of revolutions is n = 2.6.

Solution:

The radius of the turntable is,

$\begin{gathered} r=\frac{D}{2} \\ r=\frac{30.5}{2} \\ r=15.25\text{ cm} \\ r=0.1525\text{ m} \end{gathered}$

The moment of inertia of turntable is,

$\begin{gathered} I=\frac{1}{2}mr^2 \\ I=\frac{1}{2}\times0.25\times(0.1525)^2 \\ I=0.003kgm^2 \end{gathered}$

The angular displacement of the turntable is,

$\begin{gathered} \theta=2\pi n \\ \theta=2\pi\times2.6 \\ \theta=16.34\text{ rad} \end{gathered}$

By the kinematics equation, the angular acceleration of the turntable in terms of angular velocity and angular displacement is,

$\begin{gathered} \omega^2_2-\omega^2_1=2\alpha\theta \\ \alpha=\frac{\omega^2_2-\omega^2_1}{2\theta} \end{gathered}$

Substituting the known values,

$\begin{gathered} \alpha=\frac{3.49^2-0}{2\times16.34} \\ \alpha=0.373rads^{-2} \end{gathered}$

Thus, the torque acting on the turntable is,

$\begin{gathered} \tau=I\alpha \\ \tau=0.003\times0.373 \\ \tau=0.001119\text{ Nm} \\ \tau=11.19\times10^{-4}\text{ Nm} \end{gathered}$

Thus, the torque acting on the turntable is,

$11.19\times10^{-4}\text{ Nm}$