The total kinetic energy of a body of mass 10 | vedclass

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Question

Given total $\mathrm{KE}(\mathrm{E})=32.8 \mathrm{J},$ mass $(\mathrm{m})=10 \mathrm{kg}$

velocity $(v)=2 m / s .$ Radius $(r)=0.5 \mathrm{m}$

N

Vedclass · Accepted Answer

$  0.4$

Vedclass · Answer

Given total $\mathrm{KE}(\mathrm{E})=32.8 \mathrm{J},$ mass $(\mathrm{m})=10 \mathrm{kg}$

velocity $(v)=2 m / s .$ Radius $(r)=0.5 \mathrm{m}$

NOW

Total $\mathrm{KE}(\mathrm{E})=\mathrm{E}_{	ext {transalational }}+\mathrm{E}_{	ext {rotational }}$

$\Rightarrow E=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$

$\Rightarrow \mathrm{E}=\frac{1}{2} m v^{2}+\frac{1}{2} m k^{2}\left(\frac{v}{r}ight)^{2},$ where $\mathrm{k}=$ radius of gyration

$\Rightarrow 32.8=\frac{1}{2} 	imes 10 	imes 2^{2}+\frac{1}{2} 	imes 10 	imes k^{2}\left(\frac{2}{0.5}ight)^{2}$

$\Rightarrow 32.8=20+80 	imes k^{2}$

$\Rightarrow k^{2}=\frac{12.8}{80}=0.16$

$\Rightarrow k=0.4 m$

The total kinetic energy of a body of mass $10\ kg$ and radius $0.5\ m$ moving with a velocity of $2\ m/s$ without slipping is $32.8\ joule$. The radius of gyration of the body is .......... $m$

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