Question

Derive rate of change of angular momentum is equal to the torque ${\displaystyle \frac{d\overrightarrow{L}}{dt}}=\tau$

Answer 1

Hint
In the given question, we have to relate the torque acting on a body to the rate of change of its angular momentum. The angular momentum of a body is simply the moment of momentum, that is, the turning effect of the momentum of the body and torque is the rotating effect of force when applied to a hinged body. Having established the basic meaning of these terms, we can come to the derivation part of our solution.
$\overrightarrow{p}=m\overrightarrow{v}$ , $\overrightarrow{F_{net}}=m\overrightarrow{a}$ , $\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$

Complete step by step answer
As discussed above, the angular momentum is the moment of the linear momentum so
$\Rightarrow \overrightarrow{p}=m\overrightarrow{v}$ where $\overrightarrow{p}$ is the linear momentum, $m$ is the mass and $\overrightarrow{v}$ is the velocity vector.
The angular momentum can now be given as the cross product of the radius vector and the linear momentum, that is
$\Rightarrow \overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$ where $\overrightarrow{r}$ is the radius vector
Substituting the value of the linear momentum, we have
$\begin{array}{rl}& \overrightarrow{L}=\overrightarrow{r}\times \left(m\overrightarrow{v}\right)\\ & \Rightarrow \overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{v})\end{array}$
Taking the first derivative of the left and the right hand sides of the above equation, we get
$\Rightarrow {\displaystyle \frac{d\overrightarrow{L}}{dt}}=m[\overrightarrow{r}\times {\displaystyle \frac{d\overrightarrow{v}}{dt}}+{\displaystyle \frac{d\overrightarrow{r}}{dt}}\times \overrightarrow{v}]$
Now we all know that the rate of change of the velocity is known as acceleration and the rate of change of the radius or the distance vector is the velocity of the object. Also, the cross product of any vector with itself always yields a zero as the angle between them is zero.
We can further write our equation as
$\begin{array}{rl}& {\displaystyle \frac{d\overrightarrow{L}}{dt}}=m[\overrightarrow{r}\times \overrightarrow{a}+\overrightarrow{v}\times \overrightarrow{v}]\\ & \Rightarrow {\displaystyle \frac{d\overrightarrow{L}}{dt}}=m(\overrightarrow{r}\times \overrightarrow{a})[\because \overrightarrow{v}\times \overrightarrow{v}=0]\\ & \Rightarrow {\displaystyle \frac{d\overrightarrow{L}}{dt}}=\overrightarrow{r}\times m\overrightarrow{a}\end{array}$
From newton’s second law of motion, we know that force on a body is the product of its mass and acceleration, that is $\overrightarrow{F_{net}}=m\overrightarrow{a}$
We can use this relation in our angular momentum expression. We’ll get
${\displaystyle \frac{d\overrightarrow{L}}{dt}}=\overrightarrow{r}\times \overrightarrow{F_{net}}=\sum \overrightarrow{r}\times \overrightarrow{F}$
The summation of the products of the radius and the force is the torques acting on a body and hence we can write that
${\displaystyle \frac{d\overrightarrow{L}}{dt}}=\overrightarrow{\tau }$ where $\overrightarrow{\tau }$ is the net torque acting on the body.
We have thus successfully derived our required equation.

Note
In case you want to know the value of a term in rotational motion without doing the calculation or the derivation, we have a shortcut method for you. You just need to be aware of the laws of linear motion and then you can simply replace the terms with the analogous terms from the rotational motion. For example, we know that the rate of change of linear momentum of a body is the force acting on it. The analogous term of linear momentum is angular momentum and the analogous term for force is torque. Replace the terms and you’ll have your expression.