1.

A thin circular loop of radius R rotates about its vertical diameter with an angular frequency co. Show that a small bead on the wire remains at its lowermost point for omegale sqrt(g//R) . What is the angle made by the radius vector joining the centre to the bead with vertical downward direction for omega=sqrt(2g//R) 1 Neglect friction.

Answer» <html><body><p></p>Solution :Consider the free-body diagram of the <a href="https://interviewquestions.tuteehub.com/tag/bead-394814" style="font-weight:bold;" target="_blank" title="Click to know more about BEAD">BEAD</a> when the radius vector <a href="https://interviewquestions.tuteehub.com/tag/joining-525969" style="font-weight:bold;" target="_blank" title="Click to know more about JOINING">JOINING</a> the centre of the wire makes an angle `theta` with the vertical downward direction.We have mg `= N cos theta` and `m <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> <a href="https://interviewquestions.tuteehub.com/tag/sin-1208945" style="font-weight:bold;" target="_blank" title="Click to know more about SIN">SIN</a> thetaomega^2 = N sin theta` .These equations give` cos theta = g//Romega^2`.Since `cos theta lt= 1,` the bead remains at its lowemost point `omega lt= sqrt(g/R )`<br/> For `omega = sqrt((2g)/(R ), cos theta = 1/2 i.e theta = <a href="https://interviewquestions.tuteehub.com/tag/60-328817" style="font-weight:bold;" target="_blank" title="Click to know more about 60">60</a>^@`</body></html>


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