SATHEE: Laws Of Motion Question 390

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 $x^{2} = a y$ . 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) $μ a$

B) $μ^{2} a$

C) $\frac{1}{4} μ^{2} a$

D) $\frac{1}{2} μ a$

Show Answer

Answer:

Correct Answer: C

Solution:

For the sliding not to occur when $= \frac{1}{2} (\frac{4 M}{3}) R^{2} - [\frac{1}{2} (\frac{M}{3}) {(\frac{R}{2})}^{2} + {(\frac{R}{2})}^{2}] = \frac{13}{24} M R^{2}$

$\tan θ = \frac{d y}{d x} = \frac{2 x}{a} = \frac{2 \sqrt{a y}}{a} = 2 \sqrt{\frac{y}{a}}$

$∴ 2 \sqrt{\frac{y}{a}} \leq μ$ or $y \leq \frac{a μ^{2}}{4}$

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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 isx2=ay . If the coefficient of friction is M1 the highest distance above the x-axis at which the particle will be in equilibrium is

Options:

Answer:

Solution:

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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 $x^{2} = a y$ . If the coefficient of friction is $M_{1}$ the highest distance above the x-axis at which the particle will be in equilibrium is