A bead of mass M is constrained to slide along a thin circul

A bead of mass M is constrained to slide along a thin, circular hoop of radius I that rotates with constant anglarvelocity horizontal plane about a point on its rim aashown: (a) Find legrango s oq for motion of the bead (b) Show thai the bead oscillates like a pendulum about a pt.on the nm diametrically opposite the point about whihc the hooprotates. (c) what is the effective length of this pendulum?

Solution

bead makes an angle (theta-Q) with the verticle angle (fy-o) changes as bead rotates about z-axis

angular velocilt is w=do/dt,from now read it o(d),similarly dr/dt read as r(d)-means r dot i.e as differentiation of r

in spherical polar coordinates

K.E, T=(1/2)m{r(d)+r^2Q(d)^2+r^2sin^2Q.o(d)^2}

as there is no variation in r so  r(d)=0 and o(d)=w

T=(1/2)m{r^2Q(d)^2+r^2sin^2Q.w^2}

and P.E=v=mgrcosQ

so that langrangian is

L=T-V

=(1/2)m{r^2Q(d)^2+r^2sin^2Q.w^2}-mgrcosQ

(B) let a-radius of circle ,P is the position of bead on the wire at any instant.

as the plane of wire is rotating around the vertical with angular velocity w, we have o=wt.The bead also rotate on the wire so that Q is also as a function of time.

at any instant co ordinates of bead are

x=acoswt+acos(Q+wt)

y=asinwt+asin(Q+wt)

x(d)=-awsinwt-a(Q(d)+w)sin(Q+wt)

y(d)=awcoswt+a(Q(d)+w)cos(Q+wt)

so that K.E

T=1/2mx(d)^2+1/2my(d)^2

=ma^2/2[(Q(d)+w)^2+2(Q^2+w)cosQ+w^2]

giving dt/dQ=-ma^2w(Q(d)+w)sinQ

dt/dQ(d)=ma^2(Q(d)+w+wcosQ)

therefore Lagranges equation gives

d^2Q/dt^2=-w^2sinQ

d^2Q/dt^2+w^2sinQ=0 ifQ is small

which is the equation of S.H.M.

(c)if we compare above equation with simple pendulum then w^2=g/l

or l=g/w^2 which is the reqired length.