JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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Application +4 | -1

A solid sphere of mass

m

rolls without slipping on a horizontal surface with a linear velocity

v.

Then its total kinetic energy is

View answer and explanation

Correct option: D

\frac{{7m{v^2}}}{10}

Option D (

\frac{{7m{v^2}}}{{10}}

) is correct because the total kinetic energy of a rolling solid sphere is the sum of its translational and rotational kinetic energies.

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\mathrm{AB}

\mathrm{L}

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

x

\mathrm{A}

\mathrm{CM}

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

\mathrm{A}

\mathrm{b}

1.0 \mathrm{~g}

8 \mathrm{~cm}

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

t = 1\,sec

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