Angular momentum and its conservation provide the additional information we need to predict the motion of objects in space. It explains the arrangement of solar systems and galaxies. Here on Earth, it explains away the rotation of ocean currents (i.e. The Coriolis Effect), the changing spin of a dancer, diver or skater or the speed and direction of winds in hurricanes and tornadoes, which ultimately leads us to the swirling water in a toilet bowl!!
Angular momentum is always conserved and it is constant when there is zero (0) net torque exerted on the system. This sounds quite similar to linear momentum being constant when there is zero (0) net force exerted on the system. Therefore, we find this rotational equivalent to linear momentum as the angular equivalents of mass times velocity. That is, the moment of inertia times the angular velocity equals angular momentum.
L = Iw
Where L equals angular momentum and I equals the rotational inertia and w equals angular velocity.
This can be seen in the majestic and dazzling jumps and spins in Sasha Cohen’s short program at the Torino Olympic in 2006. This epic routine won her the silver medal in a very crowded field. In this link you will find a quantitative analysis of the basic concept but let’s look at it qualitatively below.
(a) Sasha decreases her moment of inertia, I, by bringing her mass closer to the rotation axis
(b) There are no external torques. Obviously there is an initial torque to send her into a spin but after that the Fg pulls straight down on her CoM, causing zero torque about this axis.
(c) The Fn has no effect for the same reason.
(d) There is negligible torque from her skates.
(e) Angular momentum, L, remains constant therefore, if you decrease the moment of inertia, I, you must increase angular velocity, w.
Napoleon and Deb play Tetherball at the end of the film where they hit ball in opposite directions in an attempt to wrap the ball around the pole. Once, the ball is hit, it has angular momentum around the pole. The angle between r, away from the axis, and F toward the axis is 180 and therefore , t = 0 and angular momentum is conserved. As the rope loops around the pole, its I decreases and its w increases.
But let’s look at some more exciting examples!! Here, we will introduce the concept of a point mass, such as a Phyllis jumping on a stationary merry-go-round to cause it to spin or a mouse on a turntable where I = mR^2. The odd thing to comprehend is that the Phyllis, before jumping on, has an angular momentum, with respect to the rotational axis, even though she is moving linearly. Again here, we have r and F in opposite directions, causing a net torque of zero.
Let’s take a look at some fun problems!!
Phyllis has a mass of 30 kg. She runs toward the merry-go-round at 3.0 m/s, then leaps on! The merry-go-round is initially at rest with a mass of 100 kg and a radius of 2.0 m. The merry-go-round can be considered a uniform disk, I = 1/2 mR^2. Find the angular speed of the merry-go-round after Phyllis jumps on.
A disk with a mass of .7 kg and a radius of .15 m is spinning about a vertical axis with angular speed of 9.00 rad/s. A hoop with a mass of 1.0 kg, which initially is not rotating, is dropped vertically on the disk. The two eventually come to rotate at the same speed.
(a) What is the common angular speed of the two objects after the collision?
(b) how much energy is lost to friction in the collision?
Here are a couple of awesome problems!
(1) Mario tosses .5 kg of pizza dough upward and then catches it. During each toss, the dough starts out uniformly distributed and expands. On one toss, the dough has an initial angular speed of 5.20 rad/s, starts as a 20 cm diameter pie and expands to 22 cm in diameter when it’s caught. Assume the mass of the pizza dough remains uniformly distributed. (I = 1/2 mR^2)
(a) Explain why the change in shape of the dough does change the angular speed of the dough but does not change the angular momentum of the dough.
(b) Calculate the angular speed of the dough when Mario catches it.
(2) A pigeon of mass 0.560 kg is flying horizontally at 5.0 m/s and enjoying a nice day. It suddenly runs into a stationary rod and holds on! The rod is uniform, with a length of 0.750 m and a mass of 2.00 kg and is hinged at its base. If the pigeon strikes the rod 0.25 m from the top of the rod, what is the angular velocity of the pigeon-rod system just after the collision? Irod = 1/3 mL^2
(3) A 20 kg, uniform board hangs horizontally at rest. It’s left end is connected to a support via a hinge, which is free to rotate. It is also supported by a light rope that makes a 30 degree angle to the board, and which is attached to the board a distance of 3/4 of its length from the hinge. Calculate the tension in the rope.
Here are some additional questions to keep you fresh!
1.) What is the speed of an electron in the lowest energy orbital of hydrogen, of radius equal to 5.29 x 10^-11 m? The mass of an electron is 9.11 x 10^-31 kg and its angular momentum is 1.055 x 10^-34 J s.
2.) A freely moving turntable is rotating at a steady rate when Jeffrey drops a glob of cookie dough and it attaches to the very edge of the turntable. Describe which quantities – angular velocity, angular acceleration, torque, Rotational KE, rotational inertia or angular momentum– are constant during the process. If the property changes, predict whether it increases or decreases.
3.) A 1000 kg merry-go-round is spinning, while supporting 10 acrobats, each with a mass of 50 kg. Initially the acrobats support each other to form a column very close to the axis of rotation (thus, the acrobats have no angular momentum at this location). Describe a plan to move the acrobats such that the angular velocity of the merry-go-round decreases to half its initial value.
4.) A disk of mass Is M and radius R has a rocket motor attached to its edge. Assume the rocket motor has negligible mass compared to the disk. The disk is free to rotate with negligible friction about and axis through its CoM perpendicular to the disk. The rocket motor fires, causing the disk to begin to rotate about this axis. The rocket, while it is firing, provides a constant force, Fo, tangent to the disk, for a time t. I = 1/2mR^2.
(a) Derive an expression for change in angular momentum of the disk in terms of the given quantities and physics constants.
(b) Derive an expression for the angular speed of the disk after the rocket has fired. Give the answer in terms of given quantities and fundamentals constants.
The rocket motor has now been adjusted so that it is angled to the edge of the disk, pointed inward. Assume the rocket fires for the same amount of time and applies a force of the same magnitude.
(c) how does the final angular speed of the disk compare to the answer you found in part (b)?
OK!! Here is the TEST!!
Film Clas 