Question

A uniform rod of mass $'m'$ and length $'l'$ can rotate freely on a smooth horizontal plane around a vertical axis hinged at a point $'H'$ . A point mass having the same mass $'m'$ coming with an initial speed $'u'$ perpendicular to the rod strikes the rod inelastically at its free end. Find out the angular velocity of the rod just after the collision.

Answer 1

Hint
Here we consider a uniform rod that is rotated about a horizontal plane. The rod is hinged to a particular point. The net torque at this point will be zero. We have to calculate the moment of inertia of the rod and the moment inertia of the point object. With this we can calculate the angular velocity of the rod after the collision of the rod and the particle.
The moment of inertia of a rod about an axis passing through one end and perpendicular to the length.
$I = \dfrac{{m{l^2}}}{3}$ ( Where $I$ is the moment of inertia, $m$ is the mass of the rod, $u$ is the initial speed of the object, $l$ is the length of the rod.
$L = I\omega$ (Where, $L$ stands for the angular momentum of the object, $I$ stands for the moment of inertia and $\omega$ stands for the angular velocity of the object).

Complete step by step answer

Since there is no external force is present in the horizontal plane, the angular momentum is conserved about $H$
We know that the angular momentum of an object can be written as,
$L = I\omega$ … (A)
We can write the equation, by substituting for $L$ and $I$
The angular momentum of the rod can be written as,
$L = mul$
Where $L$ is the angular momentum, $m$ is the mass of the rod, $u$ is the initial speed of the object, $l$ is the length of the rod.
The moment of inertia $I$ is the sum of the moment of inertia of the rod and the point object.
It can be written as,
$I = \dfrac{{m{l^2}}}{3} + m{l^2}$
$\dfrac{{m{l^2}}}{3}$ is the moment of inertia of the rod and $m{l^2}$ is the moment of inertia of the point object.
Substituting the values for $L$ and $I$ in equation (A), we get
$mul = \left( {\dfrac{{m{l^2}}}{3} + m{l^2}} \right)\omega$
From this, we get the angular velocity as
$\omega = \dfrac{{mul}}{{\left( {\dfrac{{m{l^2}}}{3} + m{l^2}} \right)}} = \dfrac{{mul}}{{m{l^2}\left( {\dfrac{1}{3} + 1} \right)}}$
$\Rightarrow \dfrac{u}{{\dfrac{l}{3} + l}} = \dfrac{{3u}}{{4l}}$
The angular velocity can be written as, $\omega = \dfrac{{3u}}{{4l}}$ .

Note
Moment of inertia can be defined as the tendency of a body to continue its angular motion. The radius about a given axis may be defined as the distance at which the whole mass of the body may be concentrated to have the same moment of inertia. This radius is called the radius of gyration. The moment of inertia mainly depends on the mass of the body and the axis chosen for the rotation of the body.