- Home
- Standard 11
- Physics
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 .......... $m$
$0.25$
$0.2$
$0.5$
$ 0.4$
Solution
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 $\mathrm{KE}(\mathrm{E})=\mathrm{E}_{\text {transalational }}+\mathrm{E}_{\text {rotational }}$
$\Rightarrow E=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$
$\Rightarrow \mathrm{E}=\frac{1}{2} m v^{2}+\frac{1}{2} m k^{2}\left(\frac{v}{r}\right)^{2},$ where $\mathrm{k}=$ radius of gyration
$\Rightarrow 32.8=\frac{1}{2} \times 10 \times 2^{2}+\frac{1}{2} \times 10 \times k^{2}\left(\frac{2}{0.5}\right)^{2}$
$\Rightarrow 32.8=20+80 \times k^{2}$
$\Rightarrow k^{2}=\frac{12.8}{80}=0.16$
$\Rightarrow k=0.4 m$
