Question

A 200 g, 20 cm diameter plastic disk is spun on an axle through its center by an electric motor. What torque must the motor supply to take the disk from 0 rpm to 1800 rpm in 4.0s?

Answer 1

Answer: The torque required is 0.0471 N m.

Explanation:

Mass of the disc = 200 g = 0.2 kg (1 kg =1000 g)

Radius of the disc =$\frac{Diameter}{2}$= 10 cm = 0.1 m(1 m = 100 cm)

Angular acceleration = $\alpha$

$\alpha =\frac{2\pi\times 1800}{60 \times 4 sec}=15\pi rad/s^2$

Moment of inertia = $\frac{1}{2}\times mass\times (radius)^2=\frac{1}{2}\times0.2 kg\times (0.1 m)^2=0.001 kg m^2$

$Torque=\alpha \times I=0.001 kg m^2\times 15\times 3.14 rad/s^2r=0.0471 N m$

The torque required is 0.0471 N m.