JEE Main & Advanced Physics NLM, Friction, Circular Motion Sample Paper Topic Test - Friction

  • question_answer
    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

    A) \[\mu a\]                                   

    B) \[{{\mu }^{2}}a\]

    C) \[\frac{1}{4}{{\mu }^{2}}a\]   

    D) \[\frac{1}{2}\mu a\]

    Correct Answer: C

    Solution :

    For the sliding not to occur when\[=\frac{1}{2}\left( \frac{4M}{3} \right){{R}^{2}}-\left[ \frac{1}{2}\left( \frac{M}{3} \right){{\left( \frac{R}{2} \right)}^{2}}+{{\left( \frac{R}{2} \right)}^{2}} \right]=\frac{13}{24}\,M{{R}^{2}}\]
    \[\tan \theta =\frac{dy}{dx}=\frac{2x}{a}=\frac{2\sqrt{ay}}{a}=2\sqrt{\frac{y}{a}}\]
    \[\therefore \,\,\,2\sqrt{\frac{y}{a}}\le \mu \]         or         \[y\le \frac{a{{\mu }^{2}}}{4}\]

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