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(A) The angular momentum $L$ is given by $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity. Thus,$\omega = \frac{L

Vedclass · Accepted Answer

$ \frac{3}{2}\frac{L}{md^{2}} $

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(A) The angular momentum $L$ is given by $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity. Thus,$\omega = \frac{L}{I}$.First,locate the centre of mass $(CM)$ of the system. Let the mass $m$ be at $x = 0$ and $2m$ be at $x = d$. The position of the $CM$ is $x_{cm} = \frac{m(0) + 2m(d)}{m + 2m} = \frac{2md}{3m} = \frac{2d}{3}$.The distance of mass $m$ from the $CM$ is $r_1 = \frac{2d}{3}$,and the distance of mass $2m$ from the $CM$ is $r_2 = d - \frac{2d}{3} = \frac{d}{3}$.The moment of inertia $I$ about the axis passing through the $CM$ is $I = m(r_1)^2 + 2m(r_2)^2 = m(\frac{2d}{3})^2 + 2m(\frac{d}{3})^2$.$I = m(\frac{4d^2}{9}) + 2m(\frac{d^2}{9}) = \frac{4md^2 + 2md^2}{9} = \frac{6md^2}{9} = \frac{2}{3}md^2$.Substituting $I$ into the formula for angular velocity: $\omega = \frac{L}{\frac{2}{3}md^2} = \frac{3L}{2md^2}$.

Two small balls with masses $m$ and $2m$ are attached to both ends of a rigid rod of length $d$ and negligible mass. If the angular momentum of this system is $L$ about an axis $(A)$ passing through its centre of mass and perpendicular to the rod,then the angular velocity of the system about $A$ is:

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