Question

A bead of mass m slides on a hemispherical surface with a velocity v at an angular position theta. Find the magnitude of angular momentum of the bead about o in the position shown

Answer 1

The magnitude of Angular momentum of the bead about O is-:

mvR

Answer 2

Answer:

The angular momentum of the bead is mvR.

Explanation:

Given that,

Mass of bead = m

Velocity = v

Angular position = Ф

Angular momentum :

Angular momentum is the product of the moment of inertia and angular velocity.

$L=I\omega$.....(I)

We know that,

Moment of inertia is defined as

$I= mr^2$

The angular velocity is defined as

$\omega=\dfrac{v}{r}$

Now put the value of I and ω in equation (I)

$L=mr^2\times\dfrac{v}{r}$

$L=mvr$

Angular momentum about O,

$L=\vec{r}\times \vec{mv}$

$L=mvr$

$L=mv\times R$

The angle between $m\vec{v}$ and $\vec{r}$ is 90°

So, The angular momentum is

$L =mvR$

Hence,  The angular momentum of the bead is mvR.