Assume a frictionless sphere exits in a uniform gravitational field-g . A point

Question

Assume a frictionless sphere exits in a uniform gravitational field-g . A point particle of mass m is at rest at the top of the sphere (Figure 1). This is an unstable equilibrium position. Suppose there is a tiny disturbance of the particle such that its initial speed is for all practical purposes zero and the particle heads straight down along a great circle that contains the top and bottom points of the sphere. Place the origin at the center of the circle and orient the axis system such that the positive z axis goes through the top point Part A At what angle relative to the positive z-axis does the particle just begin to lose contact with the sphere. Specify the angle in degrees gure 1of1 Submit Previous Answers Request Answer XIncorrect; One attempt remaining; Try Again Provide Feedback Next

Explanation / Answer

Let the angle be x.so,

current height of the mass = R*cos(x)

so loss in height = R*(1-cos(x))

let the velocity at this point be v.so,

increase in kinetic energy = loss in potential energy

or 0.5*mv^2 = mg*R(1-cos(x))

or v = (2*g*R*(1-cos(x))^0.5

now, at the point of losing contact, centrifugal force = mv^2/r

=(m*2*g*R*(1-cos(x)))/R

=(2*mg*(1-cos(x)))

normal reaction due to weight = mg*cos(x)

so, balancing the normal reaction and the centrifugal force,

mg*cos(x) = 2*mg*(1-cos(x))

or cos(x) = 2*(1-cos(x))

or x = 48.18 degrees

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