A thin circular wire hoop of radius R is rotated about its vertical diameter wit

Question

A thin circular wire hoop of radius R is rotated about its vertical diameter with a constant angular velocity omega. At the same time, it is displaced vertically as a given function of time, and the position of the center is given as f(t). A frictionless bead of mass m slides on the hoop. Find the differential equation of motion of the bead relative to the hoop using a cylindrical frame of reference. If f(t) =kt^2/2 where k is a constant, find the critical value of omega for stability of the bead near the bottom of the hoop. Assuming f(t) is given as in part (b) and omega^2 = 2 k + g/R find the positions of relative equilibrium of the bead on the wire and find the frequency for small oscillations of the bead near the stable equilibrium position.

Explanation / Answer

GIVEN:- AS SHOWN IN THE FIGURE AND GIVEN IN THE PROBLEM.

SOLUTION:-

(a) The radius vector joining the bead of mass 'm' and the centre of the wire hoop makes an angle of '' with the vertical as seen in the figure. As the centre of the wire hoop moves upwards vertically w.r.t. time 't' the bead tends to move downwards towards the bottom of the hoop and vice versa. This movemet of the bead can be interpreted with the help of the angle that changes w.r.t. time.

i.e. x is inversely proportional to i.e. f(t) 1/

For every change in f(t) there is small change in w.r.t. small change in time t --- i.e. f(t) 1 / d

i.e. f(t) = dx = C (1 / d) Where C is constant (ANSWER)

(b) Given --- f(t) = kt^2/2

For the bead to remain in equilibrium near the bottom of the hoop the centrifugal force acting on the bead should balance the weight of the bead i.e. m x R x Sin() x ^2 = m x g

i.e. = [g / R Sin()] ^ 0.5 (equation - 1)   

Now at the bottom of the hoop is very small = 1 degree, so Sin(1)=0.02

i.e. = [9.81 / R x 0.02]^0.5 = 22.15 / (R^0.5)

So = 22.15 / (R^0.5) (ANSWER)

(c) By referring equation - 1, For =3.14 & time t=0, 2(k+g) / R = g / R (0.055) i.e. 0.11k + 0.11g = g

i.e. k = 8.09g i.e. k = 79.4 So x = 0 (ANSWER)

For = 1 degree = 0.02 radians 2(k+g) / R = g / R (0.02) i.e. 0.04(k+g) = g i.e. k = 24

So x = C x 1/0.02 = 50 (Assume C=1) So x = 50 (ANSWER)

and So t = (2x50 / 24)^0.5 = 2.04 secs Now frequency f = 1/T where T = 2t = 2(2.04) = 4.08sec approx 4 secs

So f = 1/T = 1/4 = 0.25 hz Therefore frequency of bead f = 0.25 hertz (ANSWER)

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