Physics, 09.03.2020 23:43 thepatatolife

A sanding disk with rotational inertia 1.2 x 10⁻³ kg. m² is attached to an electric drill whose motor delivers a torque of 16 N. m. Find (a) The angular momentum and (b) the angular speed of the disk 33 ms after the motor.

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Answer from: proxydayz

The angular momentum of the sanding disk is 0.528 kg.m^2.s^-1 and its angular speed is 440 s^-1

Explanation:

Given:

Torque ( $\t$ ) delivers by the electric motor = $16\ N. m$

Time taken = 33 ms (milliseconds) = $33\times 10^-3 \ sec$

Moment of Inertia,MOI ( ) = $1.2\times 10^-\³ \ kg.m\²$

We have to find the angular momentum () and angular speed ( $\omega$ ).

Notes:

Angular momentum in terms of torque and time , $dL=dt \times \t$

And

Angular momentum in terms of MOI and speed, $L=I\times \omega$

So,

a) Angular momentum.

⇒ $dL=dt \times \t$

⇒ $dL=33\times 10^-^3\times 16$

⇒ $dL=0.528\ kg\ m^2 \ s^-^1$

The angular momentum of the sanding disk is 0.528 kg.m^2.s^-1

b)Angular speed.

⇒ $L=I\times \omega$

⇒ $\omega =\frac{L}{I}$

Plugging the values of $L= 0.528\ kg.m^2.s^-^1$ and MOI $(I) = 1.2\times 10^-^3\ kg. m^2$

⇒ $\omega = \frac{0.528\ kg\ m^2\ s^-^1}{1.2\times 10^-^3\ kg\ m^2}$

⇒ $\omega = 440\ s^-^1$

The angular speed of the disk is 440 s^-1

So the angular momentum of the disk is $0.528\ kg.m^2.s^-^1$ and the angular speed is $440\ s^-^1$ .

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