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6.System of Particles and Rotational Motion
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एक बॉल, मेज पर बिना फिसले चलती है। कुल ऊर्जा का कितना भाग घूर्णन में लगेगा ?
A$7 : 2$
B$2 : 9$
C$2 : 5$
D$2 : 7$
(AIPMT-1994) (JEE MAIN-2022)
Solution
Total kinetic energy = linear kinetic energy + rotational kinetic energy
$=\frac{1}{2} m v^{2}+\frac{1}{2} l \omega^{2}$
$I=\frac{2}{5} m r^{2}$
Total kinetic energy $=\frac{1}{2} m v^{2}+\frac{1}{2}\left(\frac{2}{5} m r^{2}\right) \frac{v^{2}}{r^{2}}$
$=\frac{1}{2} m v^{2}+\frac{1}{5} m v^{2}=\frac{7}{10} m v^{2}$
So, the ratio $=\frac{\text { Rotational } K E}{\text { Total } K E}$
$=\frac{\frac{1}{5} m v^{2}}{\frac{7}{10} m v^{2}}=2: 7$
$=\frac{1}{2} m v^{2}+\frac{1}{2} l \omega^{2}$
$I=\frac{2}{5} m r^{2}$
Total kinetic energy $=\frac{1}{2} m v^{2}+\frac{1}{2}\left(\frac{2}{5} m r^{2}\right) \frac{v^{2}}{r^{2}}$
$=\frac{1}{2} m v^{2}+\frac{1}{5} m v^{2}=\frac{7}{10} m v^{2}$
So, the ratio $=\frac{\text { Rotational } K E}{\text { Total } K E}$
$=\frac{\frac{1}{5} m v^{2}}{\frac{7}{10} m v^{2}}=2: 7$
Standard 11
Physics
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