This article will discuss on the topic of the angular momentum formula. The angular momentum formula is the rotational equivalent to the linear momentum. Both of the concepts deal with how quickly anything is moving. Moreover, it also deals with how difficult it is to change the speed.

However, linear momentum had only 2 variables that are mass and speed. The angular momentum has a starting with a very similar equation. Therefore, it might initially look to be about the same complexity. As we will see, it becomes much complicated and involves many more variables than the linear momentum.

Angular momentum is basically the product of the moment of inertia of an object and its angular velocity. Furthermore, both the quantities must be about the equal and the same axis i.e. the rotation line.

## Angular Momentum

**Angular Momentum = (moment of inertia)(angular velocity)**

L = \(I\omega\)

L = angular momentum (kg.\(m^{2}/s\)

I = moment of inertia (kg.\(m^{2}\)

\(\omega\) = angular velocity (radians/s)

**Derivation of the Angular Momentum Formula**

**We have Newton’s second law:**

\(\vec{F}\) = \(m\vec{a}\)

Now we multiply both the sides by “\(\vec{r}\) \(\times\) “, then we have

\(\vec{r}\) \(\times\) \(\vec{F}\) = \(\vec{r}\) \(\times\) m\(\vec{a}\) = \(\frac{d}{dt}\left ( \vec{r}\times \vec{p} \right )\)

(if we carry out the time derivative, the first term from the product rule is \(\vec{v}\) \(\times\) \(\vec{v}\), which of course is 0.)

This is very similar to the other form of the second law of newton:

\(\vec{F} = \frac{d}{dt}\vec{p}\),

We can read it as: a force provides us the rate of change of the momentum.

A force applied around an axis generates a rate of change of the momentum about the similar axis. This means we identify the left-hand side of the above equation with the torque \(\vec{\tau}\) and the quantity inside the brackets of the right-hand side with the angular momentum \(\vec{L} = \vec{r} \times m\vec{v}\), So we can also write the above equation as:

\(\vec{\tau } = \frac{d}{dt}\vec{L}\)

So in a sense, it’s in comparison with the second law of newton that we define this quantity like angular momentum.

**Solved Examples on Angular Momentum Formula**

#### Question:

A DVD disc having a radius of 0.0600 m, and a mass of 0.0200 kg. The moment of inertia of a disc of solid is I = \(\frac{1}{2}MR^{2}\), Where M is the mass of the disc, and R is the radius. When a DVD in a specific machine starts playing, it has an angular velocity of 160.0 radians/s. What is the angular momentum of the disc?

#### Solution:

We can find the angular momentum by using the formula, and the moment of inertia of a solid disc (ignoring the hole that is present in the middle). The angular momentum will be:

L = I\(\omega\)

L = \((\frac{1}{2}MR^{2})\omega\)

So, L = \(\frac{1}{2}(0.0200 kg)(0.0600 m)^{2}(160.0 radians/s)\)

L = 0.00576 kg.\(m^{2}\)/s

The angular momentum of this DVD will be 0.00576 kg.\(m^{2}\)/s.

#### Question:

A basketball spinning on the finger of an athlete has angular velocity \(\omega\) = 120.0 rad/s. The moment of inertia of a sphere that is hollow, where M is the mass and R is the radius. If the basketball has a weight of 0.6000 kg and has a radius of 0.1200 m, what is the angular momentum of this basketball?

#### Solution:

We can find the angular momentum of the basketball by using the moment of inertia of a sphere that is hollow, and the formula. The angular momentum will be:

L = I\(\omega\)

I = \(\frac{1}{2}MR^{2}\)

L = 0.6912kg.\(m^{2}\)/s

The angular momentum of the basketball that is spinning will be 0.6912 kg.\(m^{2}\)/s.

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Typo Error>

Speed of Light, C = 299,792,458 m/s in vacuum

So U s/b C = 3 x 10^8 m/s

Not that C = 3 x 108 m/s

to imply C = 324 m/s

A bullet is faster than 324m/s

I have realy intrested to to this topic

m=f/a correct this

M=f/g

Interesting studies

It is already correct f= ma by second newton formula…