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

If the coefficient of friction is µ, Tangent at any x distance would be => tanthetha =2x/a,

Now friction = umgcosthetha Balancing friction with mg*sin (thetha) we get, ucosthetha = sinthetha or tan (thetha) = u 2x/a = ux = au/2.

Even the highest point would be, y = (au/2)^2/a => a^2u^2/4a => au^2/4

Answer 2

tangent at any x distance would be => tanthetha =2x/a

Now

friction = umgcosthetha

Balancing friction with mg*sin(thetha) we get,

ucosthetha = sinthetha

or

tan(thetha) = u

2x/a = u

x = au/2

The highest point would be,

y = (au/2)^2/a => a^2u^2/4a => au^2/4