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A small bead of mass m is constrained to slide without friction inside a circular vertical hoop of radius r which rotates about a vertical axis at a frequency f. a. Determine the angle  where the bead will be in equilibrium – that is, where it will have no tendency to move up or down along the hoop. b. If f = 2.00 rev/s and r = 22.0 cm, what is ? c. Can the bead ride as high as the center of the circle ( = 90º)? Explain.

Answer :

Answer:

Explanation:

The bead is moving on a vertical circular path so it must have a centripetal force towards the centre.

This force is equal to m v² / r

v is velocity of bead and r is radius of the circular path.

The vertical hoop is also rotating about a vertical axis passing through the centre at frequency f so the bead will experience a cetrifugal force due to rotation of the hoop. Its value is

m ω² r . Only at the point o degree and 180 degree , these forces are opposite to each other so at these points , the bead will be in equilibrium .

mv² / r = m ω² r

v² = ω² r²

v = ω r

= 2π f r

= 2 x 3.14 x 2 x 0.22

v = 2.76 m /s

For the bead to rise upto point θ = 90 degree , height achieved is radius R

required velocity = √ 2gR

= √ 2x 9.8x.22

= 2.076 m/s

This velocity is less than the velocity calculated earlier so the bead will be able to ride the required height.

v = 2.76 m/s

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