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 1.70 revolutions, starting from rest? The turntable is a uniform disk of diameter 30.5 cm and mass 0.220 kg.

Respuesta :

ANSWER:

0.00145 J

STEP-BY-STEP EXPLANATION:

Using the rotational form of the kinematic equation to find α:

[tex]\omega^2_f=\omega^2_i+2\alpha\theta[/tex]

It started form rest:

[tex]\begin{gathered} \omega^{}_i=0\text{ rad/s} \\ \theta=1.7\text{ rev }=3.4\pi\text{rad} \end{gathered}[/tex]

Solving for α:

[tex]\begin{gathered} \alpha=\frac{\omega^2_f}{2\theta} \\ \alpha=\frac{3.49^2}{2\cdot3.4\pi} \\ \alpha=0.57rad/s^2 \end{gathered}[/tex]

The torque is given by the formula,

[tex]\begin{gathered} \tau=I\alpha \\ \text{for a uniform disk} \\ I=m\cdot\frac{r^2}{2},\text{ along the vertical a}\xi s\text{ in the center.} \\ \text{Therefore,} \\ \tau=m\cdot\frac{r^2}{2}\cdot\alpha,r=\frac{0.305}{2}=0.1525 \\ \text{ replacing} \\ \tau=0.22\cdot\frac{0.1525^2}{2}\cdot0.57 \\ \tau=0.00145\text{ J} \end{gathered}[/tex]