JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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R

is the radius of gyration of a body of mass

M

and radius

r

, then the ratio of its rotational to translational kinetic energy in the rolling condition is

View answer and explanation

Correct option: B

\frac{R^2}{r^2}

Option B (

\frac{R^2}{r^2}

) is correct because the ratio of rotational to translational kinetic energy depends on the ratio of the square of the radius of gyration to the square of the radius.

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)