| Physics 410 | Second Hour Exam | April 25, 2003 |
Instructions: Use Exam booklet, follow procedure theron concerning pledge.
Do two out of the three problems. Each part counts 16 points.
(1) Argue (giving reasons) what will be the the subsequent orbit of the wreck (the stuck-together PIVs).
I. In the possibly not-so-distant future, when most people have Personal Interplanetary Vehicles (PIVs), the dangerous game of Schicken (Space chicken) will gain popularity: two PIVs are sent on a head-on collision course, avoiding evasive maneuvers as long as possible.
Two such PIVs of equal mass are on circular orbits around Earth at Moon's radius, coasting towards each other. Alas, both engines fail, the PIVs collide and stick together.
Since the two PIV's are on circular orbits at the same radius, their speeds are equal and opposite (head-on collision course), and since their masses are the same, their total momentum is zero. It is conserved during the collision, so the (linear) momentum of the wreck vanishes, and it falls in a straight radial line toward Earth.
(2) Approximately how long (in days) does it take for the wreck to fall on the Earth? (No data needed for this, other than length of a month = 30 days)
The semi-major axis of the new orbit is 1/2 of Moon's orbit radius. By Kepler's law the period is therefore (1/2)3/2 = 1/(2Ö2) of Moon's period = 10.6 days. But the wreck hits the Earth after half its period, 5 days.
(3) The drivers eject just before the collision, with negligible change in their original velocities, but missing each other and the wreck. What is the subsequent motion of their (combined) center of mass, assuming the drivers are not rescued for a month? How long does it take their center of mass to reach the Earth?
Their center of mass moves on the same straight radial line as the wreck, but at a different speed: it is at the Earth when the drivers are on diametrically opposite points of the circle that is their orbit, 1/4 of a circle away from the collision point, and it takes 1/4 ×30 days = 7.5 days to get there.
(1) Find the angular velocity of the merry-go-round after the ball has been fired.
II. A cannon, assumed to be a point mass m, is mounted at the rim of a merry-go-round, which is a uniform disk of mass M and radius R. The system is initially at rest. Then the cannon fires a ball of mass m horizontally and in a tangential direction with speed v. Give all answers in terms of these quantities m, M, R, m and v.
The initial angular momentum (about the fixed center of the merry-go-round) was zero, the firing involved only internal forces, so the final angular momentum of the system (ball +cannon +merry-go-roud) must vanish,
|
(2) The cannon continues to fire balls at a regular rate, each with speed v with respect to the cannon. The balls are rolled up to the cannon from a supply at the merry-go-round's center. Argue that even if the rotation is frictionless, the merry-go-round's angular velocity does not continue to increase indefinitely. What is its limiting (average) value?
If the angular velocity does not change, each fired ball must have zero angular momentum (since it had zero angular momentum when it started out at the center). But it has speed v with respect to the cannon, so the cannon must have speed -v with respect to the ground, and its angular speed (as well as that of the merry-go-round, to which it is rigidly attached) is therefore v/R.
(3) But isn't that strange - the cannon recoils after each shot, hence exerts a torque, so the angular velocity must increase after each shot. So why does it subsequently decrease again? (Think of how the balls get to the cannon.)
The angular velocity decreases when the next ball is brought up to the cannon: Iw is constant, but I increases (by mR2), so w must decrease.
(1) Show that the projection of w on L (the angular momentum) is constant.
III. A rigid body of arbitrary shape is rotating about a fixed point (origin of coordinates) with angular velocity w. At first there is no external force or torque applied to the body.
The projection is w·L = 2T, the kinetic energy of rotation. If there is no torque, T is constant.
(2) Even if there is an external torque, show that the angle between the angular velocity w and the angular momentum L is always acute (less than 90° °).
Again, w·L = 2T, and this time we use the fact that kinetic energy is always positive: if the dot product is positive (sinq > 0), the angle q between the vectors is acute.
(3) Suppose that at some moment of time the rigid body's particle at coordinates (1, 0, 0) has velocity components (0, 2, 5). Find vx (the x-component of the velocity) of the particle at (0, 0, 1).
|
| Code | grade |
| 6350 | 88 |
| LARRY | 96 |
| 6982 | 77 |
| jj9405 | |
| Alekhine | 84 |
| stratified | 56 |
| 7909 | 30 |
| Bluey | 82 |
| bucinum | 78 |
| Morgan | 78 |
| avge | 74.33 |