socorraob

2021-12-31

A 250 g , 25-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 to 1900 rpm in 4.7 s ? Express your answer in newton-meters.

Jonathan Burroughs

Beginner2022-01-01Added 37 answers

Mass of the disc, m = 250 g = 0.25 kg

Diameter of the disc, d = 25 cm = 0.25 m

Disc rotate from 0 to 1900 rmp, in t = 4.7 sec.

Moment of inertia of the disc

$I=\frac{1}{2}m{\gamma}^{2}=\frac{1}{2}\left(0.25\right){\left(\frac{0.25}{2}\right)}^{2}$

$I=0.00195\text{}kg{m}^{2}$

$=0.002\text{}kg{m}^{2}$

angular frequency,$\omega =2\pi \left(\frac{1900}{60}\right)$

$\omega =198.96\text{}ra\frac{d}{s}$

$\omega =\alpha$

$\alpha =\frac{\omega}{t}=\frac{198.96}{4.7}=42.33\text{}ra\frac{d}{{s}^{2}}$

The torque is

$\tau =I\alpha =\left(0.00195\right)\left(42.33\right)$

$=0.0825\text{}N\cdot m$

Diameter of the disc, d = 25 cm = 0.25 m

Disc rotate from 0 to 1900 rmp, in t = 4.7 sec.

Moment of inertia of the disc

angular frequency,

The torque is

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