Answer [1]: As you make you way towards the center of the carousel the moment of inertia (or rotational inertia) of the "system", i.e., you and the carousel, decreases. Since angular momentum (equal to the product of the moment of inertia and the angular velocity) must be conserved, the angular velocity increases.
Answer [2]: The Mississippi River runs from north to south, which means that mass is being transported to regions where the tangential velocity at the Earth's surface is greater. There are no external torques actng on the Earth and so to conserve angular momentum - the product of the moment of inertia and angular velocity - a slowing down of the rotation rate must occur. Therefore, the length of the day increases, albeit by an exceedingly small amount!
Answer [3]: The whole population walking from west to east would produce angular momentum that is in the same direction as that resulting from the Earth's rotation. However, since there are no external torques on the system, i.e., the Earth and its inhabitants, the additional angular momentum created by the walkers would be compensated by a reduction of the angular momentum due to the Earth's rotation. Thus, the rotation rate of the Earth would decrease, leading to an increase in the length of the day.
If they stopped walking, conservation of angular momentum means that the Earth's rotation would increase and so the day would regain its original length.
Thank goodness the population generally walks around in random directions, i.e., their walking is not correlated!
Answer [4]: When the car starts to move clockwise it acquires angular momentum; since there are no external torques on the system (consisting of the car and bicycle wheel), conservation of angular momentum means that the wheel will acquire a counter-clockwise angular velocity, producing an equal but opposite angular momentum so the total angular momentum remains zero.
When the car stops, the wheel stops also, and when the car is set in reverse, i.e., counter-clockwise, the wheel acquires a clockwise rotation to compensate. If the masses of the car and wheel are the same then the angular velocities will be equal and opposite, since equal masses at equal radii have equal moments of inertia (rotational inertia). If the car is more massive then the angular velocity of the wheel will be greater than that of the car; if the wheel is more massive then the angular velocity of the car will be greater than that of the wheel.
Answer [5]: Without the rotor by the tail, the angular momentum of the blades of the main rotor would produce an equal but opposite angular momentum of the helicopter when it takes off, i.e., the main rotor and the helicopter would rotate in opposite directions! The tail rotor produces a torque that stops the rotational motion of the helicopter.
In order to maintain stability, if the angular velocity of the main rotor increases (decreases) then the angular velocity of the tail rotor - hence, the torque it produces - will have to increase (decrease) also. If the small rotor fails then the helicopter will rotate ... likely with catastrophic results.
Answer [6]: The most likely path is "C". The reason is there are two velocity components when the ball reached the end of the rod; a radial component (along the rod, i.e., in the direction of A), and a tangential component (in the direction of D). Thus, the resultant is a velcity vector in the C direction!
Answer [7]: The total angular momentum of the Earth-Moon system has several contributing sources; from the rotation of the Earth on its axis, from the rotation of the Moon on its axis and the rotation of the Moon about the Earth. A reduction in the Earth's angular momentum about its own axis must be compensated by an increase in one of the other sources. In fact, because of the friction between the Ocean floor and the water, the tidal bulge is 'slightly ahead' of where it would be if the Earth didn't rotate, i.e., directly in-line with the Moon:
The gravitational attraction between the bulge nearer the Moon and the Moon itself tends to increase the orbital speed of the Moon, which, in turns increases the angular momentum of the Earth-Moon system. However, that's not the whole story! In order for the Moon to be in a 'stable' orbit, the Laws of physics - namely, the balance between the centripetal and gravitational forces acting on the Moon - require that the Moon moves slightly farther away from the Earth, also increasing the angular momentum of the Earth-Moon system!
In fact, the Moon is moving away from the Earth at the rate of a few centimeters (~3cm) per year. This increase is measured using laser light reflecting from a mirror left on the Moon by Apollo astronauts.
Another interesting fact, since the Earth's rotation is slowing down it means that in the past the day was shorter than it is now. Assuming the slowing rate is constant, you can work out that 500 million years ago (5 million centuries) the day was 7500s shorter, i.e., it was about 22 hours long! That means also, 500 million years ago the year was about 398 days long!
Answer [8]: This is accomplished by using a set of beveled gears. Here is a simplified picture of what happens; it shows how two shafts can rotate at different, continuously variable, speeds. The differential gearing in a car is somewhat more complicated but it serves to demonstrate the effect.
In each case shafts A and B are not connected directly to each but via a third beveled gear, C; this ensures that A and B rotate in the same direction! Assume that A is rotating at a constant angular speed. Then the tangential speed (vt=Rw, where w is the angular speed and R the radius) of the beveled gear attached to A will vary depending on the 'take-off' point of C; at the top the tangential speed will be greatest while at the bottom the tangential speed will be smallest. As a result the angular speed of C will vary depending on the 'take-off' point on A. Since C then drives B, the angular speed of C determines the angular speed of B. Clearly, by varying the take-off position, the relative angular speeds of A and B can be changed.
We can see here in this simplified pictures that there are two 'awkward' problems:
A rather more sophisticated series of gears is used to compensate for these problems.
Answer [9]: If the egg is uncooked and you touch the shell briefly while it is rotating, the fluid inside the egg continues to rotate. When you release the shell, friction between the fluid and the shell will start the (whole) egg rotating again! Of course, if the egg is cooked, there is no fluid inside the egg and so a cooked egg, once stopped, will not start again.
Answer [10]: Since the propellor is rotating counter-clockwise (to you), its angular momentum vector is directed towards the rear of the airplane.
For the nose to rise, a torque is applied to the airplane, in the direction shown (a). By Newton's 2nd Law of rotation, this torque will produce a change of angular momentum in the same direction as the torque. That means the airplane will tend to turn towards the left during you take-off (b).
Now, in level flight, if you turn to the right, the torque is directed downward. The change in angular momentum is downward, so the resultant angular momentum is tilted downward, i.e., the nose of the airplane will tend to veer upward. Of course, an experienced pilot will "feel" these effects are make the necessary adjustments.
Will a similar effect occur with a jet-powered airplane?
Answer [11]: There are several ways to answer this problem, but since these brain "busters" are about angular momentum ... that's what we'll use! The diagram below shows a vertical view of the scenario.
First, we have to explain why the can rotates when water flows out, shown in (a). We note that when the water emerges it has momentum (p), in the direction shown. Thus, each stream of emerging water has angular momentum about the rotation axis O, with the angular momentum vector directed inward. When the can is, initially, at rest, i.e., not spinning, it has zero angular momentum. But when the water emerges, and since the total angular momentum of the system must be conserved, the can and contents must rotate, about O, in such a way that their angular momentum vector is directed outward ... that direction is counter-clockwise ... so the total angular momentum reains zero.
In (b), we have the situation with the water entering the can from outside. We can see that the entering water has angular momentum about O, with its vector is now directed outward. A similar argument to that above means that when the can is released it will rotate, about O, in such a way that its angular momentum vector is directed inward ... that direction is clockwise. So, the can will rotate in a direction opposite to that in (a).
Note, if the holes are not angled in such a way that the water emerges or enters approximately tangentially, but radially, as shown below:
then the emerging (or entering) water has zero angular momentum about O. As a result, the can will not rotate in either case!
Can you figure out, in cases (a) and (b), whether the can will speed up or slow down as the water emerges or enters? Think, moment of inertia!
Answer [12]: It is true that when the object is moving in a circle, the force is perpendicular to the velocity and so it does not produce a tangential acceleration per se nor does it do work. However, when the force is increased and the orbit radius is decreased, the actual path taken by the object is shown below.
Now, you'll notice that the applied force is no longer perpendicular to the direction of motion; there is a tangential component of the force (Ft) that will increase the (linear) speed of the object! This component is effective until the object reaches its "new" circular orbit. So, during the transition from the initial orbit to the final orbit the object experiences a tangential force that increases its speed.
In summary, this force is doing work on the object during the transition (only) and in so doing it is increasing its kinetic energy. Look, here's a neat way to show that ...
However, the force is: F = mv2/r,
which we can re-write in terms of the angular mometum ( L = mvr), i.e.,
During the process of changing orbit, angular momentum is conserved, i.e., L is constant, since there is no external torque on the system because the line of the force F passes through the axis of rotation. Therefore,
where K is the kinetic energy of the object. So, the work done during the transition from one orbit to the other is the same as the increase in kinetic energy.