a thin circular loop of radius R rotates about its vertical diameter with an angular frequency omega.Show that a small bead on the wire remains at its lowermost point foromega=square root of g divided by square root of R .What is the angle made by the radius vector joining the centre to the beed with the vwetical downward direction for omega equals square root of 2 g divided by R space end root?Neglect friction
Could you explain the answer with the steps?Thank you in advance

Asked by vasugiakash | 13th Aug, 2014, 09:23: PM

Expert Answer:

Answered by Jyothi Nair | 14th Aug, 2014, 10:41: AM