Laws Of Motion Question 390
Question: A bead of mass m is located on a parabolic wire with its axis vertical and vertex directed towards downward as in figure and whose equation is$ x^{2}=ay $ . If the coefficient of friction is $ M _{1} $ the highest distance above the x-axis at which the particle will be in equilibrium is
Options:
A) $ \mu a $
B) $ {{\mu }^{2}}a $
C) $ \frac{1}{4}{{\mu }^{2}}a $
D) $ \frac{1}{2}\mu a $
Show Answer
Answer:
Correct Answer: C
Solution:
For the sliding not to occur when $ =\frac{1}{2}( \frac{4M}{3} )R^{2}-[ \frac{1}{2}( \frac{M}{3} ){{( \frac{R}{2} )}^{2}}+{{( \frac{R}{2} )}^{2}} ]=\frac{13}{24}MR^{2} $
$ \tan \theta =\frac{dy}{dx}=\frac{2x}{a}=\frac{2\sqrt{ay}}{a}=2\sqrt{\frac{y}{a}} $
$ \therefore 2\sqrt{\frac{y}{a}}\le \mu $ or $ y\le \frac{a{{\mu }^{2}}}{4} $
