The rotational kinetic energy of a solid sphe | vedclass

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Question

$mgh =\frac{1}{2} mv ^{2}+\frac{1}{2} I \omega^{2}$

$\left( I =\frac{2}{5} mR ^{2}\right)$

$v=10 m / s$

Rotational Kinetic Energy$= K _{ r }=

Vedclass · Accepted Answer

$60$

Vedclass · Answer

$mgh =\frac{1}{2} mv ^{2}+\frac{1}{2} I \omega^{2}$

$\left( I =\frac{2}{5} mR ^{2}\right)$

$v=10 m / s$

Rotational Kinetic Energy$= K _{ r }=\frac{1}{2} I \omega^{2}$

$=\frac{1}{2}\left(\frac{2}{5}\right) MR ^{2} \frac{v^{2}}{ R ^{2}}$

$K _{ r }=60 J$

The rotational kinetic energy of a solid sphere of mass $3 \;kg$ and radius $0.2\; m$ rolling down an inclined plane of height $7\; m$ is

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