4. (30 points) Consider a \"half-pipe\" (a frictionless rail in the shape of a s

Question

4. (30 points) Consider a "half-pipe" (a frictionless rail in the shape of a semi-circle of radius R) on a turntable rotating counterclockwise at w. We work in the rotating frame, and represent Shaun White by a bead of mass m which starts essentially at rest at the origin ( 0) and slides along the rail to boad aves point A ( = ). (a) Parameterizing rto-R sinoz + R(1-cos ), compute the work done by the centrifugal force from, the origin to generic point , and verify that it equals the change in centrifugal potential, (b) What is the speed (relative to the turntable) of the bead as it reaches point A? (c) What is the speed of the bead relative to the ground at the same moment? (d) Express the speed u(d), ie, the velocity as a function of . (Check: does your expression make sense at 0 and ?) (e) Find the normal force on the bead as a function of

Explanation / Answer

4. for the given half pipe on the turn table

a. r(phi) = Rsin(phi) i + R(1 - cos(phi))j ( where i and j are unit vectors along x and y directions respectively)

at any point phi

centrifugal force Fc = mw^2*r(phi)

hence work done by centrifugal force = Fc.dr

now, dr = Rcos(phi)d(phi)i + R(sin(phi))d(phi)j

hence

dW = mw^2*(Rsin(phi) i + R(1 - cos(phi))j).( Rcos(phi)d(phi)i + R(sin(phi))d(phi)j)

dW = mw^2*R^2(sin(phi)cos(phi) + sin(phi) - sin(phi)cos(phi))d(phi)

dW = mw^2*R^2*sin(phi)d(phi)

net work, integrating from phi = 0 to phi = phi

W = mw^2*R^2*(1 - cos(phi))

npw, |r|^2 = r.r = R^2sin^2(phi) + R^2(1 - cos(phi))^2 = R^2(1 + 1 - 2cos(phi)) = 2R^2(1 - cos(phi))

hence W = mw^2|r^2|/2

b. speed relative to turn table = dr/dt

v = dr/dt = Rcos(phi)d(phi)/dt i + R(0 + sin(phi))d(phi)/dt j = R(cos(phi)*phi'i + sin(phi)phi'j)

at point A, phi = 180 deg

v = R(-phi'i)

c. speed relative to the ground = v + w x r(phi) = -R*phi' i + w k x (Rsin(phi) i + R(1 - cos(phi))j)

Vr = v + w x r(phi) = -R*phi' i + wRsin(phi) j + wR(1 - cos(phi))i

Vr = R(-phi' + w(1 - cos(phi)))i + wRsin(phi)j

for phi = 180 deg

Vr = R(-phi' + 2w)i

d. v(phi) = R(cos(phi)*phi'i + sin(phi)phi'j)

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