Rolling Motion of a Rigid Body:
Please refer to Figure 10.31 and 10.32 on page 290 (Serway). The body shown is rolling without slipping. Consequently, point P (which is momentarily in contact with the ground) is not moving relative to the ground. Or,
Other points on the body have various velocities. Since point P is the center of rotation, Serway has drawn radii from Q and Q' to show that the velocities of Q and Q' are perpendicular to those radii. Likewise, P' has a tangential velocity perpendicular to its radius (not shown). Since the body is momentarily rotating about P with a certain angular velocity omega, and the body is rigid, all points in the body have the same angular velocity. The radius of the body is the distance from P to CM; P to P' is twice this radius. Thus the tangential velocity of point P' is twice that of CM:
Since the body is rotating about point P, its kinetic energy is
Serway then substitutes the parallel axis theorem to obtain
Thus, you may compute the kinetic energy of a rolling body by using either of the formulas above. The second formula may be interpreted as the sum of the translational kinetic energy (2nd term) and the rotational kinetic energy about the center of mass (1st term).
Consider a body rolling down an incline. Our previous analyses always assumed sliding (no rolling), so no energy went into making the body rotate. When rotational energy is considered, we would expect the acceleration to be less.
Example 1: A comparison of bodies moving on an incline
Angular Momentum (optional):
The definition (for a particle) is
The magnitude of the angular momentum is rp times the sine of the angle between r and p. It is often easier to compute the magnitude of the angular momentum and assign a direction by inspection, rather than by doing a cross product. For example, a point mass m moving in a circle of radius r with tangential velocity v is easy: mvr is the magnitude of the angular momentum (the sine of 90o is one).
For a rigid body (a collection of an infinite number of particles), a sum (integration) must be used. This results in
You're probably wondering why we're suddenly dealing only with the z-direction. While the angular momentum and the angular velocity are vectors, I is a tensor which in general has nine components. For our work, we deal only with rigid body rotation about an axis of symmetry, and can treat I as if it were a scalar. (You don't need to know this!)
Example 2: The Earth's angular momentum (optional)
Example 3: Using the cross product to obtain angular momentum (optional)
Example 4: The angular momentum of a composite body (optional)
Net Force and Torque from Momentum:
For linear motion, the net force is the time rate of change of the linear momentum. For rotational motion, the net torque equals the time rate of change of the angular momentum:
(These are relativistically valid whereas the usual sum of the forces = ma form is not.)
Example 5: Acceleration using the derivative of momentum
Conservation of Angular Momentum (all refer to Serway's 3rd Edition):
Did you ever wonder why, as a skater draws her arms and legs inward, she speeds up? Maybe you've tried the experiment shown in Figure 11.17 on page 320. Have you looked at Example 11.15 on page 321? All of these exhibit the principle of conservation of angular momentum!
Please look at Example 11.13 on page 320. As the student walks in toward the center, his distance (r) from the center changes. If we approximate him as a point mass, his angular momentum changes from mviri to mvfvf. So the turntable has to change its angular momentum to compensate. Just as when applying conservation of linear momentum, where external forces could not be present (or we'd have to redefine the system), here we can't have external torques, such as friction. The system is the student and turntable; the forces between his feet and the table are therefore internal to the system. If friction at the bearings of the table were present, however, it would be an external force (and cause an external torque).
The student in Conceptual Example 11.14 on page 320 pulls his arms inward, thus reducing his moment of inertia. Since that would change his angular momentum, the angular velocity has to increase.
For Example 11.15 on page 321, you have to think in terms of vectors. The wheel is initially rotating with its (vector) angular velocity (and thus angular momentum) pointing up. When the student turns the wheel over, she has changed the angular momentum, since it now points down! That's why the chair (with her) has to rotate counterclockwise, to produce an upward momentum twice as large as the wheel's downward momentum, so that the total angular momentum is the same as before.