Answer:
After 2.0s the angular momentum is [tex]L= 2(4A+3B+2C+D)x[/tex]
Explanation:
Let us call forces acting on the rod, A, B, C, and D, and the separation between them [tex]x[/tex] .
Then, the torque due to force A is
[tex]\tau_a = 4Ax[/tex],
due to the force B
[tex]\tau_b = 3Bx[/tex],
due to force C
[tex]\tau_c = 2Cx[/tex],
and the torque due to force D is
[tex]\tau_d = Dx[/tex].
Therefore, the total torque on the the stick is
[tex]\tau_{tot} =\tau_a+\tau_b+\tau_c+\tau_d[/tex]
[tex]\tau_{tot} =4Ax+3Bx+2Cx+Dx[/tex]
[tex]\tau_{tot} =x(4A+3B+2C+D)[/tex]
Now, this torque causes angular acceleration [tex]\alpha[/tex] according to the equation
[tex]I \alpha = \tau_{tot}[/tex]
where [tex]I[/tex] is moment of inertia of the stick and it has the value
[tex]I = \dfrac{1}{3} m(4x)^2[/tex]
Therefore the angular acceleration is
[tex]\alpha = \dfrac{\tau_{tot} }{I}[/tex]
[tex]\alpha =\dfrac{x(4A+3B+2C+D)}{\dfrac{1}{3}m(4x)^2 }[/tex]
[tex]\boxed{\alpha =\dfrac{3(4A+3B+2C+D)}{16mx } .}[/tex]
Now, the angular momentum [tex]L[/tex] of the stick is
[tex]L = I\omega[/tex],
where [tex]\omega[/tex] is the angular velocity.
Since [tex]\omega = \alpha t[/tex], we have
[tex]$L = \dfrac{1}{3}m (4x)^2 *\dfrac{3(4A+3B+2C+D)}{16mx }* t$[/tex]
[tex]L= (4A+3B+2C+D)x t[/tex]
Therefore, [tex]t = 2.0s[/tex], the angular momentum is
[tex]\boxed{ L= 2(4A+3B+2C+D)x. }[/tex]
