A bead of mass m is welded at the periphery of the smoothly pivoted disc of mass m and radius R as shown in figure. The pivot axis is horizontal and through the centre of disc. The disc is free to rotate in vertical plane. The disc is released .......?

Question

A bead of mass #m# is welded at the periphery of the smoothly pivoted disc of mass m and radius #R# as shown in figure. The pivot axis is horizontal and through the centre of disc. The disc is free to rotate in vertical plane. The disc is released from rest from position shown. Find the speed of bead at its lowest position?

A) #sqrt{(3/8) gR}#
B) #sqrt{(3/4) gR}#
C) #sqrt{(4/3) gR}#
D) #sqrt{(8/3) gR}#

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Answer 1 · 2017-04-02T10:33:31

D)

Relatively to the lowest path position, the set has a potential energy given by

#V=2Rmg#

and the kinetic energy at the lowest position is

#T=1/2 m v^2+1/2J_0 (dot theta)^2# where

#J_0 = 1/2 m R^2# is the uniform mass disk moment of inertia concerning the rotation pivot.

We know also that #v = dot theta R# now equating energies

#2Rmg = 1/2mv^2+1/2(1/2m R^2)v^2/R^2# and after simplifications

#v = sqrt(8/3Rg)# or option D)