Question

Derive an expression that rel
momentum with the angular velocity of rigid body.​

Answer 1

Explanation:

For the rigid body, the angular momentum L is the product of the momentum of inertia and the angular

velocity : L=Iw.

For a point of mass , angular momentum can be expressed as the product of linear momentum and the radius .

( r ): L = mvr

Answer 2

Question:

Derive an expression that relates angular momentum with the angular velocity of rigid body.

Solution:

See the attachment:

Let's consider a rotating body having n numbers of particles at different radius.

Let their masses be $m_1,m_2,m_3,m_4, - - - - m_n$ and radii at which they are located be $r_1,r_2,r_3,- - - - r_n$

Now, we know that:

• $\vec{L} = \vec{r} \times \vec{p}$ where L is angular momentum, r is radius, p is linear momentum.
• It can also be written as $L= rp sin \theta$
• Here, we will take the angle b/w r and p as 90° as we are considering the particle to be perpendicular to the plane it is rotating on. [sin90=1]
• $L= r \times mr \omega \implies L= mr^2 \omega$

Calculating L for each n particle of the rotating body.

For particle m1:

• $L_1= m_1(r_1)^2 \omega$

For particle m2

• $L_2= m_2(r_2)^2 \omega$

Similarly, for particle mn

• $L_n=m_n(r_n)^2 \omega$

Sum total angular momentum:

$L=L_1+L_2+L_3+ - - - - - - + L_n \\\\ L= m_1(r_1)^2 \omega+m_2(r_2)^2 \omega+ - - - - +m_n(r_n)^2 \omega \\\\ (Taking ~\omega~common)~we~have: \\\\ L=\bigg( m_1(r_1)^2 +m_2(r_2)^2 + - - - - +m_n(r_n)^2 \bigg) \omega \\\\ L = \bigg(\sum_{(i=1)}^{(n)} m_i(r_i)^2 \bigg) \omega \\\\ But, here ~we~know~ that~ \sum_{(i=1)}^{(n)} m_i(r_i)^2~is~moment~of~inertia. \\\\ \therefore \vec{L} = I \times \vec {\omega} \: \: \: [\omega= angular ~velocity.]$

The relation here we got is :

Angular momentum = Moment of Inertia x Angular velocity.