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In the non-inertial frame of a car we have two forces, gravity and the inertial force, which combine to give an ‘effective gravity’. If A~ is the accelaration of the car with respect to the ground:
geff = g − A The important point is that the buoyant force directly opposes this effective gravity. In the case of a helium balloon, the buoyant force is greater than the effective gravitational force, so the displacement of the balloon will be in the direction of the buoyant force, upward and forward. The equilibrium angle is the same as the angle of the effective gravity:
θeq = arctan
A
g
An additional (and unneccessary) complication, the above treatment of the buoyant force depends on the car being enclosed. If the car is not enclosed, the buoyant force will align with the true gravitional field. In that case the helium balloon will tilt backwards, even ignoring air resistance.
a)
The centrifugal force always points radially outward from the axis of rotation. Near the North pole that direction is south. The coriolis force on someone increasing their radius will be against the direction of rotation. Since the earth rotates to the east, the coriolis force in this case is westward.
b)
East is the direction of rotation, so the person on the equator is moving tangen- tially in the direction of rotational motion. The centrifugal force is, as always, radially outward, which from the perspective of someone standing on the equa- tor is directly up. The coriolis force is also radially outward (or up since at the equator) here. An object moving in the direction of rotational motion requires a stronger centripetal acceleration to maintain circular motion. That increase in net centripetal force is produced by a reduction in the normal force from the ground. This is felt as an effective force in the up direction.
c)
The centrifugal force is independent of motion, so as above the centrifugal force is upward. In this case there is no coriolis force, however. This can be seen easily in the formula for the coriolis force Fcor = 2mr˙ × Ω. If one is walking south at the equator, r˙ is antiparallel to Ω and the cross product gives zero.
The derivation of the ‘fictitious forces’ came from the expansion of the acceler- ation in (9.32):
( d^2 r d t^2
S 0
d d t
S
[(
dr d t
S
]
+ Ω ×
[(
dr d t
S
]
In the simplification that follows Ω is assumed constant, therefore its time derivative is zero. Relaxing that assumption gives us the additional term:
( d^2 r d t^2
S 0
= (previous) + Ω˙ × r
Which is then carried through to equation (9.34) where it appears as a third fictitious force (the change in order in the cross product contains the negative sign picked up in this manipulation):
Faz = mr × Ω˙
Problem 5
First consider the problem without including the rotational motion. In this case we have projectile motion as described below:
z(t) = v 0 t −
g t^2
From this we can also get that the time of impact is ti = 2v 0 /g and the
maximum height is h = 12 v 02 g.^ We want to consider the effect of rotational motion on the east-west motion of the projectile. This motion will be impacted only by the coriolis force, the centrifugal force is radially outward and does not act in the tangential, or east-west, direction. Our coordinate system for our location (in the northern hermisphere) is arranged tangentially to the ground, with z up, x the east-west axis with east positive, and y the north-south axis with north positive. This is a right-handed coordinate system. To a first approximation our projectile motion holds for the z direction. However, accounting for the coriolis force gives:
Fcor = 2mr˙ × Ω = 2m (v 0 − g t) ˆz × Ω = 2m (v 0 − g t) (−ˆx) cos λ
The cos λ comes from the cross product, remember that the earth’s latitude is measured from the equator, not from the pole as we normally do in physics. Now
We find that the Ω which balances the tangential forces is independent of r!
c)
Given that there is no intial velocity, and that the forces are all balanced with this choice of Ω, the mass stays at rest indefinitely.
d)
Since the centrifugal force is completely balanced by the gravitational force, the only force at work is the coriolis force. Consider a set of axes in the vicinity of a point (r, φ) on this parabolic dish, the ˆt axis pointing tangentially up the surface, and the φˆ axis pointing in the rotational direction. We can write down the coriolis force in these axes,
Fcor = 2m( ˙tˆt + φ˙ˆφ) × Ω = − 2 mΩcos(θ) ˙tˆφ + 2mΩsin(θ) φ˙ˆt
Remember that θ is just the angle defined above. From this we can write down our differential equations describing the motion.
φ¨ = −2Ωcos(θ) ˙t ¨t = 2Ωsin(θ) φ˙
We can differentiate each equation, and then plug the other on into it, in order to get an ordinary differential equation. This gives us, after an integration of the result of the previously described manipulation,
φ¨ = −4Ω^2 (sinθ cosθ)φ + Cφ ¨t = −4Ω^2 (sinθ cosθ)t + Ct
However, now the motion is clear. These two equations describe harmonic oscillation with identical periods. The other important thing to note is that the amplitudes of the motion must be the same in each direction, as is clear from our original differential equations in which time derivatives in one equation are directly proportional to the other. So this is not just elliptical motion, it is circular motion. A full solution would incorporate the initial conditions to fix all unknown constants, but a qualitative description of the motion is sufficient for the problem as stated.
