A 2 kg mass is rotating on a circular path of radius 0.8 m with angular velocity of 44 rad/sec. If radius of the path becomes 1 m, then what will be the value of angular velocity?

A 2 kg mass is rotating on a circular path of radius 0.8 m with angula

1.	A 2 kg mass is rotating on a circular path of radius 0.8 m with angular velocity of 44 rad/sec. If radius of the path becomes 1 m, then what will be the value of angular velocity?
Answer» <html><body><p>28.16 rad/sec <br/>19.28 rad/sec<br/>8.12 rad/sec <br/>35.26 rad/sec </p>Solution :Mass (m) = 2 <a href="https://interviewquestions.tuteehub.com/tag/kg-1063886" style="font-weight:bold;" target="_blank" title="Click to know more about KG">KG</a> , <br/> initial radius of the path `(r_(1)) = 0.8` m , initial angular velocity (`omega_(1))` = 44 rad/sec and <a href="https://interviewquestions.tuteehub.com/tag/final-461168" style="font-weight:bold;" target="_blank" title="Click to know more about FINAL">FINAL</a> radius of the path `(r_(2)) = 1` m <br/> Initial moment of inertia , `I_(1) = m r_(1)^(2) = 2 xx (0.8)^(2) = 1.28 kg m^(2)` <br/> and final moment of inertia , `I_(2) = mr_(2)^(2) = 2 xx (1)^(2) = 2 kg m^(2)` <br/> From the law of conservation of angular <a href="https://interviewquestions.tuteehub.com/tag/momentum-1100588" style="font-weight:bold;" target="_blank" title="Click to know more about MOMENTUM">MOMENTUM</a> <br/> `I_(1) omega_(1) = I_(2) omega_(2) or omega_(2) = (I_(1) omega_(1))/(I_(2)) = (1.28 xx 44)/(2) = 28.16` rad/sec</body></html>