Question

An LP turntable must spin at 3.500 rad/s to play a record. How much torque must the motor deliver if the turntable is to reach its final angular speed in 1.800 revolutions, starting from rest? The turntable is a uniform disk of diameter 30.80 cm and mass 0.2500 kg.

Answer 1

Answer:

0.00321 Nm

Explanation:

$\omega_f$ = Final angular velocity = 3.5 rad/s

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = 2 rev

Equation of rotational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{3.5^2-0^2}{2\times 2\pi \times 1.8}\\\Rightarrow \alpha=0.54156\ rad/s^2$

Moment of inertia is given by

$I=mr^2\\\Rightarrow I=0.25\times 0.154^2\\\Rightarrow I=0.005929\ kgm^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=0.005929\times 0.54156\\\Rightarrow \tau=0.00321\ Nm$

The torque required is 0.00321 Nm