Assignment 8 Solutions

1. (Problem 9.11 ***)   (a) The KE evaluated in the inertial frame So is T = 1/2m vo2 = 1/2m (v + W×r)2, so the Lagrangian is L = 1/2m (r· + W×r)2 - U.

(b) The derivatives of L are as follows:

L
x
=
m (
×
r
 
 + W×r) ·
x
 (W×r) - U
x
=
m (
×
r
 
 + W×r) ·(0,Wz, -Wy) - U
x
=
m[(
×
r
 
 + W×r) ×W]x - U
x
 .
We can combine this with the two corresponding y and z equations, into a single vector equation

L
r
 = m
×
r
 
 ×W + m (W×r) ×W+ F
Similarly

d
dt
 L
×
r
 
 = m(
××
r
 
 + W×
×
r
 
 )
So the three Lagrange equations are mr¨ = F + 2 m (r· ×W) + m(W×r) ×W, in agreement with Eq.(9.34).

2. (Problem 9.14 **)    In the rotating frame of the bucket, the water is in equilibrium and its surface is an equipotential surface for the combined gravitational force (PE = mgz) and centrifugal force (force = m W2 r and hence PE = -m W2 r2/2). Therefore, the surface is given by mgz - m W2 r2/2 = const, or z = [(W2 r2)/2 g] + const, which is a parabola, as claimed.

3. (Problem 9.29 **)   (a) With vx o = vy o = 0 and vz o = vo, the trajectory given by (9.73) is
x = -Wsinq(vo- [1/3] g t)t2,     y = 0, and     z = vot - 1/2g t2.
We see that the ball does not stray in the north-south (y) direction (at least to first order in W), but it does move in the east-west (x) direction. The time for the ball to return to the ground is found from the z equation to be tgr = 2 vo/g. Substituting this value into the x equation, we find xgr = -(4 Wvo  3 sinq)/(3 g2). Since this is negative, we conclude that the Coriolis effect causes the ball to land to the west of its point of departure.

(b) The westerly displacement is maximum at the equator, where sinq = 1, and the displacement is

4  W vo  3
3  g2
 = 4 ×(7.3 ×10-5) ×(40)3
3 ×(9.8)2
 » 0.065 m
or about 6 cm - not an effect that would be easy to detect.
(c) On the upward journey, the Coriolis force accelerates the ball to the west and on the downward journey to the east. Thus vx starts from zero, increases (to the west) as the ball climbs, and decreases back to zero by the time the ball returns to the ground. Throughout the trip vx is to the west, and the ball lands to the west of its starting position. The dropped ball starts with vx = 0 at the top and its whole journey is downward, so that the Coriolis force accelerates it to the east throughout. Thus vx is to the east at all times, and the ball lands to the east of its initial position. In the pictures, the Coriolis effect is much exaggerated.

4. (Problem 1.46 **)   (a) As seen in the inertial frame S the puck moves in a straight line with f = 0 and r = R - vot

(b) As seen in S¢, r¢ = r = R - vot and f¢ = f- wt = - wt. This path is sketched on the right. Initially, the puck moves inward with speed vo but also downward with speed wR. It curves to its right, passing through the center and continuing to curve to the right until it slides off the turntable.
(c) In polar coordinates we have r = vt,   f = wt hence x = vt coswt,   y = vt sinwt. x is an odd function of t, and y is even, therefore all the odd powers of x in an expansion of y must vanish. Also y = 0 when x = 0 since the particle goes through the origin, so a = 0.
To get y(x) we neet to express t in terms of x, that is, solve the first equation for t. We do this approximately for small t, for example by iteration: to lowest order, coswt = 1 hence vt » x. To quadratic order, coswt » 1 - 1/2(wt)2 = 1 - 1/2 (wx/v)2 so ct » x + 1/2(w/v)2 x3. Substitute this into the equation for y, also expanded in powers of wt to find
y = (w/v)x2 + (7w3/6v3) x4 + ...

5. w must be perpendicular to the horizontal plane, that is, along the local vertical. The total angular velocity of the pendulum's plane of oscillation with respect to an inertial frame is of course W+ w. In order that this total vector cause no Coriolis force (in the horizontal plane where the pendulum bob is constrained to move) it must itself be in the horizontal plane. Thus the picture looks a shown on the right, and clearly the magnitude is w = Wcosq. This is also the local vertical component of Ω, but the direction is opposite - the rotating observer R thinks everything rotates opposite to the way the inertial observer I "knows" R to rotate.