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=(D)/(2) \\ r=(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=(1)/(2)mr^2 \\ I=(1)/(2)*0.25*(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*2.6 \\ \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=(\omega^2_2-\omega^2_1)/(2\theta) \end{gathered}`

Substituting the known values,

`\begin{gathered} \alpha=(3.49^2-0)/(2*16.34) \\ \alpha=0.373rads^(-2) \end{gathered}`

Thus, the torque acting on the turntable is,

`\begin{gathered} \tau=I\alpha \\ \tau=0.003*0.373 \\ \tau=0.001119\text{ Nm} \\ \tau=11.19*10^(-4)\text{ Nm} \end{gathered}`

Thus, the torque acting on the turntable is,

`11.19*10^(-4)\text{ Nm}`