Respuesta :
Answer:
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 ( [tex]\t[/tex]) delivers by the electric motor = [tex]16\ N. m[/tex]
Time taken = 33 ms (milliseconds) = [tex]33\times 10^-3 \ sec[/tex]
Moment of Inertia,MOI ( [tex]I[/tex]) = [tex]1.2\times 10^-\³ \ kg.m\²[/tex]
We have to find the angular momentum ([tex]L[/tex]) and angular speed ([tex]\omega[/tex]).
Notes:
Angular momentum in terms of torque and time , [tex]dL=dt \times \t[/tex]
And
Angular momentum in terms of MOI and speed, [tex]L=I\times \omega[/tex]
So,
a) Angular momentum.
⇒ [tex]dL=dt \times \t[/tex]
⇒ [tex]dL=33\times 10^-^3\times 16[/tex]
⇒ [tex]dL=0.528\ kg\ m^2 \ s^-^1[/tex]
The angular momentum of the sanding disk is 0.528 kg.m^2.s^-1
b)Angular speed.
⇒ [tex]L=I\times \omega[/tex]
⇒ [tex]\omega =\frac{L}{I}[/tex]
Plugging the values of [tex]L= 0.528\ kg.m^2.s^-^1[/tex] and MOI [tex](I) = 1.2\times 10^-^3\ kg. m^2[/tex]
⇒ [tex]\omega = \frac{0.528\ kg\ m^2\ s^-^1}{1.2\times 10^-^3\ kg\ m^2}[/tex]
⇒ [tex]\omega = 440\ s^-^1[/tex]
The angular speed of the disk is 440 s^-1
So the angular momentum of the disk is [tex]0.528\ kg.m^2.s^-^1[/tex] and the angular speed is [tex]440\ s^-^1[/tex] .
