Assume a particle has angular velocity w about a pivot point. We define the angular momentum L of the particle about the point as L = r´p, where r is the displacement vector of the particle from the pivot point and p is its momentum. The direction of L is perpendicular to both r and p. Let r and p lie in the x-y plane, as shown in the figure below.
Then L = rpsinqk. For a particle moving in a circular path sinq=1, L = rpk = rmv k = mr2w k = Iw, since w = wk.
L = Iw.
The change in angular momentum of the particle is
.
The last term on the right is proportional to v´v and therefore is zero. We have
.
The result,
t
= dL/dt,is the rotational analog of Newton's second law.
The angular momentum of a rigid object rotating about one of its symmetry axes is the sum of the angular momenta of all its parts. It is a measure of an object's rotational motion about this axis. The angular momentum L then is the product of the object's moment of inertia I times its angular velocity w about the chosen axis.
L = Iw
Angular momentum is a vector. The direction of the angular momentum of a rigid object rotating about one of its symmetry axes is the direction of the angular velocity (given by the right hand rule). The change DL in the angular momentum of an object is equal to the angular impulse tDt. You give an object an angular impulse by letting a torque act on it for a time interval Dt.
angular impulse = torque ´ time
D
L = tDtThe total angular momentum of a system of objects is conserved if no external torque acts on the system. The total angular momentum of the universe about any axis is therefore conserved. The angular momentum of a single object, however, changes when a net torque acts on the object for a finite time interval. Conversely, if no net torque acts on an object, its angular momentum is constant.
If no external torque acts on a system of interacting objects, then their total angular momentum is conserved.
In the video clip shown below the total angular momentum of the system points upward. The person is stopping a spinning wheel and the stool starts to spin.
Some of the first few frames:
As the person applies a torque to the wheel, the wheel applies a torque to the person. The magnitudes of the angular momenta of the wheel and of the person change at the same rate, but their sum remains constant.
 | A light rod 1m in length rotates in the xy plane about a pivot through the
rod's center. Two particles of mass 4kg and 3kg are connected to its
ends. Determine the angular momentum of the system at the instant the
speed of each particle is 5m/s.
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| A 60kg woman stands at the rim of a horizontal turntable having a moment
of inertia of 500kgm2, and a radius of 2m. The turntable is
initially at rest and is free to rotate about a frictionless vertical axis
through its center. The woman then starts walking around the rim
clockwise (as viewed from above the system at a constant speed of 1.5 m/s
relative to the Earth. (a) In what direction and with what angular speed does the turntable rotate? (b) How much work does the women do to set herself and the turntable in motion?
|
Many spinning objects are not spinning about a fixed axis, but have translational as well as rotational motion. A football thrown by a quarterback, for example, is spiraling through the air. Fortunately, this type of motion is not too complicated to describe. Each object has a point about which it naturally spins. This point is the center of mass. It is the point about which the objects mass is evenly balanced, no matter what the orientation of the object.
We can always separate the complicated motion of an object into a rotation about an axis through the center of mass and a translation of the center of mass.
The kinetic energy of an object with translational as well as rotational motion is the sum of the kinetic energy of the motion of the center of mass and the kinetic energy of the motion about the center of mass.
Consider a uniform disk of radius r and mass m rolling on a flat surface in the x-direction. The displacement Dx and the angular displacement Dq are related through Dx = rDq.
The magnitudes of the linear velocity and the angular velocity are related through vCM = rw.
The kinetic energy of the disk is the sum of the kinetic energy of the motion of the center of mass (1/2)mvCM2 = (1/2)mr2w2, and the kinetic energy of the motion about the center of mass, (1/2)Iw2. The moment of inertia I of a uniform disk about an axis perpendicular to the plane of the disk through its CM is (1/2)mr2. The kinetic energy of the disk therefore is (3/4)mr2w2.
Consider a composite object, such as the two joined disks in the figure below.
The center of mass of this object is at the origin. To find the moment of inertia of the object about the CM we can use the parallel axis theorem. This theorem states that the moment of inertia of an object about any axis is equal to the sum of two terms. The first term is of the moment of inertia of the object about a parallel axes through its center of mass. The second term is the product of the mass of the object M times the square of the distance R of its center of mass from the axis in question.
I = ICM + MR2.
We can treat the composite object as the sum of its parts, and for each part calculate the moment of inertia about the z-axis.
For disk 1 we have I1 = ICM1 + M1R12, and for disk 2 we have I2 = ICM22 + M2R22.
The moment of inertia of the composite object about the z-axis then is I = I1+I2.
For a uniform disk of mass M the moment of inertia about an axis through its center of mass and perpendicular to the plane of the disk is (1/2)MR2.
For the object in the figure we therefore have for the moment of inertia about the z-axis
I = (3/2)MR2 + (3/2)MR2 = 3MR2.
Assume a particle is displaced a distance rdq along a circular path. If a tangential force acts on the particle through the displacement, then the work done by this force is W = Frdq = rFdq = tdq.
W = tdq
The rate at which work is done (i.e. the power) is P = dW/dt = tdq/dt = tw.
P = tw.
The net work done on a rotating object is equal to the change in the rotational kinetic energy of the object.
By clicking the button below, you can play or to step through a video clip frame-by-frame. In the clip the same torque acts on objects with different moments of inertia. The torque is the product of a weight and a small lever arm. The moment of inertia of the ruler-like object changes because masses are added at larger distances away from the center. When the weight has dropped through the same distance Dy, the same work has been done and the system has the same kinetic energy, since it starts from rest. Neglecting friction W = tq = mgDy = (1/2)Iw2. The torque however does not act through the same amount of time Dt, and therefore the system does not have the same angular momentum L = tDt = Iw. The angular momentum L = (2WI)1/2 is larger for the system with the larger moment of inertia I, while the angular speed w = (2W/I)1/2 is smaller for the system with the larger moment of inertia I. (IE4 required)
| A puck of mass 80g and radius 4cm slides along an air table at a speed of
1.5m/s. It makes a glancing collision with a second puck of radius 6cm
and mass 120g (initially at rest) such that their rims just touch. The
pucks stick together and spin after the collision. (a) What is the angular momentum of the system relative to the center of mass? (b) What is the angular velocity about the center of mass?
|
When you push on an object with a force directed towards its center of mass, the object will accelerate. But it will not start rotating about its center of mass. You are not applying a torque. The lever arm is zero. No torque implies no angular acceleration. When you push on it with a force not directed towards the center of mass, you exert a torque about the CM, because the force now has a lever arm. This will result in linear as well as angular acceleration of the object. The linear acceleration is a result of the force and the angular acceleration is a result of the torque.
If a ball hits a bat right at the center of mass the bat will accelerate backward without rotating. The bat's handle will jerk backward in the batter's hand. If the ball hits farther away from his hand, the bat will accelerate backward, but at the same time start rotating about its center of mass. This rotation will move the handle forward, while the translation moves it backward. If the ball hits at just the right spot, called the center of percussion, the backward and forward accelerations exactly cancel and the batter can swing the bat smoothly without feeling much of a jerk. The center of percussion of baseball bats, tennis rackets, golf clubs, and other sporting equipment is called one of their sweet spots.
Links to other Web materials:
| Rotation |
| Rotational Dynamics |
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