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Standard XII

Physics

Relation between Linear and Angular Speed

If R is the...

Question

If $R$ 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 :

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

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

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

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Solution

The correct option is C $\frac{R^{2}}{r^{2}}$
$\frac{K_{R}}{K_{T}} = \frac{\frac{1}{2} I ω^{2}}{\frac{1}{2} m v^{2}} = \frac{I ω^{2}}{m v^{2}}$

$= \frac{(m R^{2}) (v / r)^{2}}{m v^{2}} = \frac{R^{2}}{r^{2}}$

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If 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 :

The correct option is C R2r2KRKT=12Iω212mv2=Iω2mv2=(mR2)(v/r)2mv2=R2r2

If $R$ 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 :

The correct option is C $\frac{R^{2}}{r^{2}}$
$\frac{K_{R}}{K_{T}} = \frac{\frac{1}{2} I ω^{2}}{\frac{1}{2} m v^{2}} = \frac{I ω^{2}}{m v^{2}}$

$= \frac{(m R^{2}) (v / r)^{2}}{m v^{2}} = \frac{R^{2}}{r^{2}}$