JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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A thin circular disc of mass

M

and radius

R

is rotating about an axis passing through centre of mass and perpendicular to plane of disc with constant angular velocity

\omega

. The angular momentum of the disc about same axis is :

View answer and explanation

Correct option: B

\frac{1}{2} \mathrm{MR}^2 \omega

Option B (

\frac{1}{2} \mathrm{MR}^2 \omega

) is correct because it is the correct formula for the angular momentum of a thin circular disc rotating about an axis through its center and perpendicular to the plane.

More Questions from SYSTEM OF PARTICLES

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It is easier for a swimmer jumping into water from a height to describe a loop in the air by A uniform thin rod

\mathrm{AB}

of length

\mathrm{L}

has linear mass density

\mu(x)=a+\frac{b x}{L}

, where

x

is measured from

\mathrm{A}

. If the

\mathrm{CM}

of the rod lies at a distance of

\left(\frac{7}{12}\right) \mathrm{L}

from

\mathrm{A}

, then a and

\mathrm{b}

1.0 \mathrm{~g}

is rotating in a circle of diameter

8 \mathrm{~cm}

. Its moment of inertia will be The angular velocity of a particle is given by

\omega = {t^4} + 2{t^3}

then its angular acceleration of the particle at

t = 1\,sec

\left( {rad\,{s^{ - 2}}} \right)