Question

The figure above represents a stick of uniform density that is attached to a pivot at the right end and has equally spaced marks along its length. Any one or a combination of the four forces shown can be exerted on the stick as indicated. All four forces are exerted on the stick that is initially at rest. What is the angular momentum of the stick after 2.0s ?

Answer 1

Answer:

After 2.0s the  angular momentum is $L= 2(4A+3B+2C+D)x$

Explanation:

Let us call forces acting on the rod, A, B, C, and D, and the separation between them $x$ .

Then, the  torque due to force A is

$\tau_a = 4Ax$,

due to the force B

$\tau_b = 3Bx$,

due to force C

$\tau_c = 2Cx$,

and the torque due to force D is

$\tau_d = Dx$.

Therefore, the total torque on the the stick is

$\tau_{tot} =\tau_a+\tau_b+\tau_c+\tau_d$

$\tau_{tot} =4Ax+3Bx+2Cx+Dx$

$\tau_{tot} =x(4A+3B+2C+D)$

Now, this torque causes angular acceleration $\alpha$ according to the equation

$I \alpha = \tau_{tot}$

where $I$ is moment of inertia of the stick and it has the value

$I = \dfrac{1}{3} m(4x)^2$

Therefore the angular acceleration is

$\alpha = \dfrac{\tau_{tot} }{I}$

$\alpha =\dfrac{x(4A+3B+2C+D)}{\dfrac{1}{3}m(4x)^2 }$

$\boxed{\alpha =\dfrac{3(4A+3B+2C+D)}{16mx } .}$

Now, the angular momentum $L$ of the stick is

$L = I\omega$,

where $\omega$ is the angular velocity.

Since $\omega = \alpha t$, we have

$L = \dfrac{1}{3}m (4x)^2 *\dfrac{3(4A+3B+2C+D)}{16mx }* t$

$L= (4A+3B+2C+D)x t$

Therefore,   $t = 2.0s$, the angular momentum is

$\boxed{ L= 2(4A+3B+2C+D)x. }$