JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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Application +4 | -1

A loop of radius 3 meter and weighs

150 \mathrm{~kg}

. It rolls along a horizontal floor so that its centre of mass has a speed of 15

\mathrm{cm} / \mathrm{sec}

. How much work has to be done to stop it

View answer and explanation

Correct option: A

3.375 \mathrm{~J}

Option A (3.375 J) is correct because the work done to stop the loop is equal to its total kinetic energy, considering both translational and rotational motion.

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)