Homework 1 Flashcards
the surface of the ocean is an equipotential surfac
To prove this, consider a drop of sea water on the surface of the ocean.
The drop is in equilibrium (relative to the earth’s reference frame) under the influence
of three forces: the earth’s gravitational pull mg, the tidal force Ftid, and the pressure
force Fp of the surrounding sea water. Since a static fluid cannot exert any shearing
force, the pressure force F p must be normal to the surface of the ocean. (It is in fact just
the buoyant force of Archimedes’ principle.) Since the drop of water is in equilibrium,
it follows that mg + Ftid must likewise be normal to the surface.
Euler’s theorem
the most general motion of any body relative to a fixed point
0 is a rotation about some axis through 0. It
implies that to specify a rotation about a given point 0, we need give only the direction
of the axis about which the rotation occurred and the angle through which the body
rotated.
Velocity vector in terms of omega and r
v=omega x r.
It is perhaps worth emphasizing that there is a corresponding relation for
any vector fixed in the rotating body. For example, if e is a unit vector fixed in the
body, then its rate of change, as seen from the non-rotating frame,
de/dt = omega x e. An important result.
Addition of Angular Velocites
A final basic property of angular velocities that is worth mentioning is that relative
angular velocities add in the same way as relative translational velocities. We know
(in the framework of classical mechanics) that if two frames 2 and 1 have relative
velocity v21 and if a body 3 has velocity v 32 relative to frame 2, then the velocity of 3
relative to frame 1 is just the sum
V31 = V32 + V21.
Suppose instead that frame 2 is rotating with angular velocity co 21 relative to frame
1 (both frames with the same origin 0) and that body 3 is rotating (about 0) with
angular velocities &31 and 0)32 relative to frames 1 and 2. Now consider any point
r fixed in body 3. Its translational velocities relative to frames 1 and 2 must satisfy
(9.24). According to (9.22) this means that
0)31 x r = (0)32 x r) + (co21 x r) = (0)32 + 0)21 ) x r
and, since this must be true for any r, it follows that
(-1)31 = (1)32 te)21.
Time Derivatives in a Rotating Frame, Angular Velocity of Earth
We are now ready to consider the equations of motion for an object that is viewed
from a frame S that is rotating with angular velocity S2 relative to an inertial frame
So. While our conclusions will apply to any rotating frame, by far the most important
example is a frame attached to the rotating earth, and this is the example that you can
keep in mind. This being the case, let us pause to calculate the angular velocity of the
earth, which rotates on its axis once every 24 hours. 5 Therefore, for a frame attached
to the earth Omega = 7.3 x 10-5 rad/s. There are other noninertial
effects (notably the tides), associated with the orbital motion of the earth and moon.
Rates of Change in different frames
rate of change of vector Q relative to inertial frame, rate of change of same vector Q relative to rotating frame. Have subscripts
Relation between rates of change of vector in inertial and noninertial frame
dQ/dt = dQ/dt + OmegacrossQ. Apply this iteratively to get the result of Newton’s second law in rotating frame
9.6
Let h (8) denote the height of the ocean at any point T on the surface, where h (8) is measured up from the level at the point Q of Figure 9.5 and 6 is the polar angle TOR of T . Given that the surface of the ocean is an equipotential, show that h(8) = ho cost 8, where ho = 3 Mn, R:/ (2 Me clo3). Sketch and describe the shape of the ocean's surface, bearing in mind that ho << Re . [Hint: You will need to evaluate Utid ( T ) as given by (9.13), with d equal to the distance MT . To do this you need to find d by the law of cosines and then approximate d -1 using the binomial approximation, being very careful to keep all terms through order (Re /do) 2 . Neglect any effects of the sun.]
9.11
In this problem you will prove the equation of motion (9.34) for a rotating frame using the
Lagrangian approach. As usual, the Lagrangian method is in many ways easier than the Newtonian
(except that it calls for some slightly tricky vector gymnastics), but is perhaps less insightful. Let S be
a noninertial frame rotating with constant angular velocity St relative to the inertial frame S o . Let both frames have the same origin, 0 = 0’. (a) Find the Lagrangian L = T — U in terms of the coordinates
r and r of S. [Remember that you must first evaluate T in the inertial frame. In this connection, recall
that vo = v + St x r.] (b) Show that the three Lagrange equations reproduce (9.34) precisely.
9.14
I am spinning a bucket of water about its vertical axis with angular velocity C2. Show that,
once the water has settled in equilibrium (relative to the bucket), its surface will be a parabola. (Use
cylindrical polar coordinates and remember that the surface is an equipotential under the combined
effects of the gravitational and centrifugal forces.)
Complications Tidal
Although the theory just presented is basically correct, especially in the middle of
the larger oceans, the real situation involves many intriguing complications. Perhaps
the most important complication is the effect of the continental land masses. So far I
have pretended that the oceans cover the whole world, allowing the tidal forces of the
moon and sun to collect the two bulges of Figure 9.5. But the presence of continents
can affect our conclusions, leading sometimes to smaller and sometimes to larger tides.
A small sea, such as the Black Sea or even the Mediteranean, that is shut off from the
main oceans by land will obviously exhibit much smaller tides than we have calculated. here. On the other hand, the tides moving across a large ocean can be blocked by the
bordering continents and can build up to much greater heights. And tides entering
narrow tapering estuaries can cause quite dramatic “tidal bores.”
Moon Alone, Sun Alone
54, 25 cm
Types of Tides
Spring Tides, Neap Tides, High Tides, Low Tides
gravitational constant
g = G*Mearth/RearthRearth
Tidal Potential and Other Expressions
Ueg (P) — Ue g (Q) = Utid(Q) — Utid(P), Ueg (P) — Ueg(Q) = mgh.
Utid = minus G *Mmoon test massreciprocal of distance from test charge to moon plus horizontal distance from center of earth divided by distance to moon squared
