Question

obtain an expression for the angular momentum of a body rotating with uniform angular velocity​

Answer 1

Answer:

L=I ω

Explanation:

Answer 2

Angular Momentum of a body:

• Let's consider a rigid body rotating with a constant angular velocity of $\omega$ about specific geometry axis.

• The body will be composed of 'n' number of particles each having a distinct mass (m1, m2, m3, . . . .) and rotating at specific distance ( r1, r2, r3, . . .) from the axis.

• The individual angular momentum of the particles will be (m1)(r1)²$\omega$ and so on. . .

Magnitude of the total angular momentum will be algebraic addition of angular momentum of all the particles because the vector is in the same direction :

$L = m_{1} { (r_{1}) }^{2} \omega +m_{2} { (r_{2}) }^{2} \omega + . \: . \: .m_{n} { (r_{n}) }^{2} \omega$

$\implies L = \bigg \{m_{1} { (r_{1}) }^{2}+m_{2} { (r_{2}) }^{2} + . \: . \: .m_{n} { (r_{n}) }^{2} \bigg \} \omega$

$\implies L = I \times \omega$

[ So, expression derived ]