Answer:
[tex]-20.0 kg m^2/s[/tex]
Explanation:
The angular momentum of an object in rotation is given by
[tex]L=I \omega[/tex]
where
I is the moment of inertia
[tex]\omega[/tex] is the angular speed
In this problem, initially we have
[tex]I=2 kg m^2[/tex] is the moment of inertia of the wheel
[tex]\omega_i = 6.0 rad/s[/tex] is the initial angular speed
So the initial angular momentum is
[tex]L_i = I\omega_i = (2)(6.0)=12 kg m^2/s[/tex]
Later, a counterclockwise torque of
[tex]\tau=-5.0 Nm[/tex] is applied
So the angular acceleration of the wheel is:
[tex]\alpha = \frac{\tau}{I}=\frac{-5.0}{2}=-2.5 rad/s^2[/tex] in the direction opposite to the initial rotation.
As a result, the final angular velocity of the wheel will be:
[tex]\omega_f = \omega_i + \alpha t[/tex]
where
t = 4.0 is the time interval
Solving,
[tex]\omega_f = +6.0 +(-2.5)(4.0)=-4 rad/s[/tex]
which means that now the wheel is rotating in the counterclockwise direction.
Therefore, the new angular momentum of the wheel is:
[tex]L_f = I\omega_f =(2)(-4.0)=-8.0 kg m^2/s[/tex]
So, the change in angular momentum is:
[tex]\Delta L=L_f - L_i = (-8.0-(12))=-20.0 kg m/s^2[/tex]
