A cylinder of radius R has the y axis as its central axis, and therefore appears

Question

A cylinder of radius R has the y axis as its central axis, and therefore appears as a circle in the diagram below. The cylinder does not move. A massless string of total length b (greater than TR) is attached to the cylinder as shown in the diagram. A point mass m is attached to the lower end of the string. The motion is in the xz plane. The point P is defined so that the string above P is in contact with the cylinder, and the string elow P is not in contact with the cylinder, as shown in the diagram. This means the ocation of P on the string changes as m swings left and right. We will use the angle ? as our generalized coordinate for this problem (-t/2

Explanation / Answer

From the given diagram

length of string = b

a. radius of circle = R

hence length og string in contact with the cylinder = l

l = pi/2*R - theta*R = R(pi/2 - theta)

hence

length of string not in contact with the cylinder = b - l = b - R(pi/2 - theta)

b. x coordinate = Rcos(theta) + (b - l)cos(90 - theta) = Rcos(theta) + R(pi/2 - theta)sin(theta)

z coordinate = Rsin(theta) - (b - l)sin(90 - theta) = Rsin(theta) - R(pi/2 - theta)cos(theta)

c. Langragian

T = 0.5mv^2

whwere m is mass of particle and v its velcity given by

and

V = mg(z) = mg( Rsin(theta) - R(pi/2 - theta)cos(theta))

hence Langragain can be written as

L = T - V = 0.5mv^2 - mg( Rsin(theta) - R(pi/2 - theta)cos(theta))

d. from langrage's equations of the second kind

-ve of gradient of potential is m*a ( a being acceleration)

hence

mg(Rcos(theta) - R(0 - 1)(-sin(theta))) = ma

a = gR(cos(theta) - sin(theta))

e. for time independent quantity, total energy is conserved

TE = KE + PE = 0.5mv^2 - mg( Rsin(theta) - R(pi/2 - theta)cos(theta))

f. d(TE)/dt = 0

mv*a = mgR(cos(theta) - sin(theta))w

now w = v/( b - R(pi/2 - theta))

m*a = mgR(cos(theta) - sin(theta))/( b - R(pi/2 - theta))

dv/dt = wdv/d(theta)

w = v/( b - R(pi/2 - theta))

dv = ( b - R(pi/2 - theta))*dw

w( b - R(pi/2 - theta))*dw = gR(cos(theta) - sin(theta))d(theta)/( b - R(pi/2 - theta))

integrating

integral(w( b - R(pi/2 - theta))*dw) from (w = 0 to w = w0) =integral( gR(cos(theta) - sin(theta))d(theta)/( b - R(pi/2 - theta from (theta = thetao to theta = theta)

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