A barbell is mounted on a nearly frictionless axle through its center (Figure 11.105). At this instant, there are two forces of equal magnitude applied to the system as shown, with the directions indicated, and at this instant, the angular velocity is 60 Rad/s, counterclockwise. In the next $\mathbf{0}\mathbf{.}\mathbf{001}\mathbf{\text{s}}$,the angular momentum relative to the center increases by an amount 2.5 kg.meter square per second. What is the magnitude of each force? What is the net force?

Momentum is the product of the mass and the velocity of the object. Any object moving with mass possesses momentum. The only difference in angular momentum is that it deals with rotating or spinning objects.

The torque applied on the object is defined as the change in the angular momentum per unit time.

$\tau =\frac{dL}{dt}$

$\tau =\frac{\Delta L}{\Delta t}$ ….. (1)

Here, L is the angular momentum.

Below figure shows the direction of the forces acting on the bare ball system.

Here, $F$is the force acting on the system of two system of two balls, $m_1$and $m_2$are masses of the two balls, and d is the distance between the two balls.

The torque acting on the system due to the constant force $\left(F\right)$is defined as the product of the force componentand $\left(F\sin \theta \right)$the perpendicular distance from the other end$\left(\frac{d}{2}\right)$.

$\begin{array}{rcl}\tau & =& \left(F\sin \theta \right)\left(\frac{d}{2}\right)\\ & =& \frac{1}{2}Fd\sin \theta \\ & & \end{array}$

Substitute $\frac{1}{2}Fd\sin \theta$for$\tau$into the equation (1) and solve for F.

\(\frac{{\Delta L}}{{\Delta t}} = \frac{1}{2}Fd\sin \theta \)

\(F = \left( {\frac{2}{{d\sin \theta }}} \right)\left( {\frac{{\Delta L}}{{\Delta t}}} \right)\) ….. (2)

Here,

The distance, $d=12\text{cm}$$d=12\text{cm}$

The angular momentum, 2.5 kg.meter square per second

The change in time, $\Delta t=0.001\text{s}$

Substitute these values into equation (2), and you have

$\begin{array}{rcl}F& =& \left(\frac{2}{\left(12\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(\sin 30°\right)}\right)\left(\frac{2.5\times {10}^{-4}}{0.001\text{s}}\right)\\ & =& 8.34\text{N}\\ & & \end{array}$

Hence, the magnitude of the force acting on the each force is $8.34\text{N}$.

The two forces are equal in magnitude, but opposite in direction. Hence, their net force is equal to zero.