Question

4 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 x² = ay. If the coefficient of friction is µ, the highest distance above
the x-axis at which the particle will be in equilibrium is
(a) µa (b) µ²a (c)1/4 µ^2 g (d)1/2µ^ g

Answer 1

Answer on Question #63996-Physics-Mechanics-Relativity

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 $\mu$ , the highest distance above the $x$ -axis at which the particle will be in equilibrium is

(a) $\mu a$ (b) $\mu^2 a$ (c) $1 / 4\mu^{\wedge}2g(d)1 / 2\mu^{\wedge}g$

Solution

Tangent at any $x$ distance would be

$$\tan \theta = y ^ {\prime} = \frac {2 x}{a}$$

The friction is

$$F _ {f r} = \mu m g c o s \theta$$

Balancing friction with $mgsin(\theta)$ we get,

$$\mu \cos \theta = \sin (\theta) \rightarrow \tan \theta = \mu$$

So,

$$\frac {2 x}{a} = \mu$$ $$x = \frac {a \mu}{2}$$

The highest point would be,

$$y = \frac {\left(\frac {a \mu}{2}\right) ^ {2}}{a} = \frac {a \mu^ {2}}{4}.$$

Answer: $\frac{a\mu^2}{4}$ .

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