# How do I calculate the angular velocity of a falling object?

## I have a problem where a falling object hits the end of a beam (which is pivoted in the centre), and sticks to it. I need to calculate the instantaneous angular velocity of the end of the beam after this collision. The object can be treated as a particle. Air resistance can be ignored. I have the distance it falls, and its mass. I also have the length of the beam, and its mass. To solve this I need the angular velocity at the very bottom of the object's motion.

Feb 24, 2016

Instantaneous angular velocity$= \frac{6 m \sqrt{2 g h}}{l \left({m}_{b} + 3 m\right)}$radian ${\text{s}}^{-} 1$

#### Explanation:

By definition angular velocity $\omega$ is for an object moving in a circular path and is connected by the expression
$v = r \omega$, .......(1)
where $v$ is linear velocity, $r$ is the radius of circle in which the object moves.

In the question we have a falling object under gravity. This is linear motion.
Assuming the object falls through a height $h$, with initial velocity being zero.
Change in potential energy $= m g h$

Once the object hits the beam which is pivoted at the centre and gets attached to it.
Let velocity of object just before the collision be equal to $v$. This can be found from the kinetic energy of the object. Using Law of conservation of Energy,
Change in $K E = \frac{1}{2} m {v}^{2} = m g h$
This gives us, ignoring air resistance
$v = \sqrt{2 g h}$, $g$ being acceleration due to gravity and $= 9.8 m {s}^{-} 2$

Let $l \mathmr{and} {m}_{b}$ be length and mass of the beam respectively. When the object collides with the beam and sticks to it

Initially, the angular momentum of the object-beam system has only contribution from the object. The beam is not moving.

$L = r m v$,
where $r$ is the radius of rotation and $v$ is the perpendicular component of velocity.

${L}_{i} = \frac{l}{2} m \sqrt{2 g h}$

After the collision, the angular velocity of the object and beam is the same. They have a combined moment of inertia of
$I = \frac{1}{12} {m}_{b} {l}^{2} + m {l}^{2} / 4$ about the pivot. The final momentum is given by:

${L}_{f} = I \omega$

$= \left(\frac{1}{12} {m}_{b} {l}^{2} + m {l}^{2} / 4\right) \omega$

As the angular momentum is conserved $\therefore {L}_{i} = {L}_{f}$.

$\frac{l}{2} m \sqrt{2 g h} = \left(\frac{1}{12} {m}_{b} {l}^{2} + m {l}^{2} / 4\right) \omega$

$\omega = \frac{\frac{l}{2} m \sqrt{2 g h}}{\left(\frac{1}{12} {m}_{b} {l}^{2} + m {l}^{2} / 4\right)}$

$= \frac{6 m \sqrt{2 g h}}{l \left({m}_{b} + 3 m\right)}$

Feb 24, 2016

$\omega = \frac{6 m \sqrt{2 g h}}{\left({m}_{b} + 3 m\right) l}$

#### Explanation:

From the conservation of energy, we can calculate the speed of the mass just before it strikes the beam.

$\frac{1}{2} m {v}^{2} = m g h$

$v = \sqrt{2 g h}$

Since we do not know the force applied by the pivot, we consider the total angular momentum about the pivot. The angular momentum of the system is conserved, as the only external force creating significant impulse in this short duration of time (by the pivot) has zero torque due to zero perpendicular distance.

Initially, the angular momentum of the mass-beam system has only contribution from the mass. The beam is not moving.

${L}_{i} = m v \frac{l}{2}$

$= m \frac{l}{2} \sqrt{2 g h}$

Finally, the angular velocity of the mass and beam is the same. They have a combined moment of inertia of $I = \frac{1}{12} {m}_{b} {l}^{2} + m {\left(\frac{l}{2}\right)}^{2}$ about the pivot. The final momentum is given by:

${L}_{f} = I \omega$

$= \left(\frac{1}{12} {m}_{b} {l}^{2} + \frac{1}{4} m {l}^{2}\right) \omega$

Since angular momentum is conserved, ${L}_{i} = {L}_{f}$.

$m \frac{l}{2} \sqrt{2 g h} = \left(\frac{1}{12} {m}_{b} {l}^{2} + \frac{1}{4} m {l}^{2}\right) \omega$

$\omega = \frac{m \frac{l}{2} \sqrt{2 g h}}{\frac{1}{12} {m}_{b} {l}^{2} + \frac{1}{4} m {l}^{2}}$

$= \frac{6 m \sqrt{2 g h}}{\left({m}_{b} + 3 m\right) l}$