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  • 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 loop remains at its lowermost point for Omega less than or equal to root of g/R . What is the angle made by the

Q5.40 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 loop remains at its lowermost point for \omega \leq \sqrt{g/R}. What is the angle made by the radius vector joining the centre to the bead with the vertically downward direction for \omega = \sqrt{2g/R}? Neglect friction.

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The FBD of the loop is given below :

                                                Rotational motion, 20343

Using equilibrium conditions we can write :

                                               mg\ =\ N\cos \Theta

and                                       ml \omega ^2\ =\ N\sin \Theta

Consider \Delta OPQ, we have :

                                                \sin \Theta\ =\ \frac{l}{R}

or                                              l\ =\ R\ \sin \Theta

Using l in above equation we get :

                                           m(R \sin \Theta) \omega ^2\ =\ N\sin \Theta

or                                                mR \omega ^2\ =\ N

And                                      mg\ =\ (mR \omega^2)\cos \Theta

Thus                                 \cos \Theta \ =\ \frac{g}{R \omega ^2}

Since                                  \omega\ =\ \sqrt{\frac{2g}{r}}

Substituting the value of \omega in the above equation:-  

                                        \frac{2g}{R}\ =\ \frac{g}{R \cos \Theta}

Thus                           \cos \Theta \ =\ \frac{1}{2}

or                                       \Theta \ =\ 60^{\circ}

Posted by

Devendra Khairwa

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