Answer:
(a) Angular momentum of disk is [tex]1.343\, \frac{kg.m^{2}}{s}[/tex]
(b) Angular velocity of the disk is [tex]353\frac{rad}{s}[/tex]
Explanation:
Given
Rotational inertia of the disk , [tex]I=3.8\times 10^{-3}kg.m^{2}[/tex]
Torque delivered by the motor , [tex]\tau =17N.m[/tex]
Torque is applied for duration , [tex]\Delta t=79ms=0.079s[/tex]
(a)
Magnitude of angular momentum of the disk = Angular impulse produced by the torque
[tex]\therefore L=\tau \Delta t=17\times 0.079\frac{kg.m^{2}}{s}[/tex]
=>[tex]L=1.343\, \frac{kg.m^{2}}{s}[/tex]
Thus angular momentum of disk is [tex]1.343\, \frac{kg.m^{2}}{s}[/tex]
(b)
Since Angular momentum , [tex]L=I\omega[/tex]
where [tex]\omega[/tex]= Angular velocity of the disk
[tex]=>\omega =\frac{L}{I}=\frac{1.343}{3.8\times 10^{-3}}\frac{rad}{s}[/tex]
[tex]\therefore \omega =353\frac{rad}{s}[/tex]
Thus angular velocity of the disk is [tex]353\frac{rad}{s}[/tex]
