Question

Derive an expression for angular momentum.

Answer 1

Hint The product of the angular velocity and the moment of inertia of an object is known as angular momentum. Furthermore, both the quantities must be about the equal and the same axis i.e. the rotation line. So by using newton’s second law concept we will be able to derive the expression for it.

Formula used
Force,
$\vec F = m\vec a$
Here,
$F$, will be the force
$m$, will be the mass
$a$, will be the acceleration

Complete Step By Step Solution
As we know from the second law of newton:
$\vec F = m\vec a$
So now we will multiply both the sides by$\vec r$, we get
$ \Rightarrow \vec r \times \vec F = \vec r \times m\vec a$
And also the above equation can be written in the form of a derivative as
$ \Rightarrow \dfrac{d}{{dt}}\left( {\vec r \times \vec p} \right)$
Along these lines, if we complete the time derivative, the initial term from the product rule $\vec v \times \vec v$ , which obviously will be zero.
This is fundamentally the same as the other type of the second law of newton:
$ \Rightarrow \vec F = \dfrac{d}{{dt}}\vec p$
We can read it as a force that provides us the rate of change of the momentum.
A force applied around a pivot produces a pace of progress of the energy about the comparable hub. This implies we recognize the left-hand side of the above condition with the torque $\vec \tau $ and the amount inside the sections of the right-hand side with the angular momentum,
$ \Rightarrow \vec L = \vec r \times m\vec v$
So we can likewise compose the above condition as:
$ \Rightarrow \vec \tau = \dfrac{d}{{dt}}\vec L$
So it might be said, it's in examination with the second law of newton that we characterize this amount like angular momentum

Note Angular momentum is the rotational equivalent of linear momentum. Just as linear momentum equals the product of mass and velocity, angular momentum equals the product of angular mass (that is, a moment of inertia) and angular velocity.
Thus, angular momentum $ = $ Moment of inertia $ \times $ Angular velocity.
Thus, if we know the linear terms of motion and if we know the radius of the circular trajectory, then the angular terms of motion can be easily determined. Radius connects the expression of angular quantities and their corresponding linear quantities.