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A ring of mass m and radius r rotates about an axis passing through its centre and perpendicular to its plane with angular velocity omega. Its kinetic energy is |
Answer» <html><body><p>`1/2mr^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)<a href="https://interviewquestions.tuteehub.com/tag/omega-585625" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGA">OMEGA</a>^(2)`<br/>`mromega^(2)`<br/>`mr^(2)omega^(2)`<br/>`1/2mromega^(2)`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/kinetic-533291" style="font-weight:bold;" target="_blank" title="Click to know more about KINETIC">KINETIC</a> energy =`1/2Iomega^(2)`, and for ring `I=mr^(2)`. Hence KE `=1/2mr^(2)omega^(2)`</body></html> | |