Answer
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Hint: To calculate the relation between moment of inertia and angular momentum, we have to know about the formula of angular momentum and moment of inertia in terms of mass, velocity i.e.,
Angular momentum $L = mvr$
Angular velocity $\omega = \dfrac{v}{r}$
Complete step by step solution:
Angular momentum is the rotational equivalent of linear momentum i.e.,
Angular momentum L $ = $ radius (r) $ \times $ linear momentum (p) …..(1)
Linear momentum p is the product of mass and velocity i.e.,
$p = m \times v$ …..(2)
From equation (1) and (2)
$L = mvr$
Now dividing and multiplying by r
$L = mvr \times \dfrac{r}{r}$
$L = (m{r^2})\left( {\dfrac{v}{r}} \right)$ …..(3)
We know that angular velocity
$\omega = \dfrac{v}{r}$ …..(4)
From equation (3) & (4)
$L = (m{r^2})\omega $
Here $m{r^2}$ is known as the moment of inertia I.
So, $\boxed{L = I\omega }$
Above expression shows the relation between moment of inertia and angular momentum.
According to above formula we can say that moment of inertia I is a quantity expressing a body’s tendency to resist angular acceleration which is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation i.e., $\boxed{I = m{r^2}}$
Unit $ \to kg{m^2}$
Dimension $ \to [M{L^2}]$
It is a linear quantity.
Note: To establish the relation between any 2 physical quantities we must know about the definitions of them and some basic relation related to them.
Angular momentum $L = mvr$
Angular velocity $\omega = \dfrac{v}{r}$
Complete step by step solution:
Angular momentum is the rotational equivalent of linear momentum i.e.,
Angular momentum L $ = $ radius (r) $ \times $ linear momentum (p) …..(1)
Linear momentum p is the product of mass and velocity i.e.,
$p = m \times v$ …..(2)
From equation (1) and (2)
$L = mvr$
Now dividing and multiplying by r
$L = mvr \times \dfrac{r}{r}$
$L = (m{r^2})\left( {\dfrac{v}{r}} \right)$ …..(3)
We know that angular velocity
$\omega = \dfrac{v}{r}$ …..(4)
From equation (3) & (4)
$L = (m{r^2})\omega $
Here $m{r^2}$ is known as the moment of inertia I.
So, $\boxed{L = I\omega }$
Above expression shows the relation between moment of inertia and angular momentum.
According to above formula we can say that moment of inertia I is a quantity expressing a body’s tendency to resist angular acceleration which is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation i.e., $\boxed{I = m{r^2}}$
Unit $ \to kg{m^2}$
Dimension $ \to [M{L^2}]$
It is a linear quantity.
Note: To establish the relation between any 2 physical quantities we must know about the definitions of them and some basic relation related to them.
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