Taylor 4.1 (today) and 4.2-4.3 (Wednesday).
For a system of n particles the kinetic energy is the sum of the individual kinetic energies:
We can follow the same argument used to derive the work -- kinetic energy theorem for a single particle to derive it for a system of particles. I will omit the (more complex) derivation here, but let's look at the kinetic energy for a system a particles, because the theorem only works if all the total work and kinetic energy are accounted for.
Now if we express vi in terms of the CM velocities, we can break the kinetic energy into a piece due to the motion of the CM and a piece due to the relative motions of the particles.
Inserting this into the kinetic energy expression, and noting that the remaining dot product (third term) sums to zero, we get:
In words, the total kinetic energy of a system is equal to the kinetic energy due to the motion of the CM plus the kinetic energy due to motion of particles of the system relative to their CM. This latter part can come from rotation of the system (a collective motion of all the particles of the system), or the motion of individual particles of the system.
If the system of particles are part of a rigid body, then all the relative velocities, v'i, are due to the object's rotation about the CM and can be related to the angular velocity and distance from the axis of rotation (passing through the CM), v'i = ωr⊥i:
relation | translational motion | rotational motion |
---|---|---|
momentum | px = mvx | Lz = Izω |
kinetic energy | T = 1/2 mv² | T = 1/2 Izω² |
equation of motion | Fx = dpx/dt | Nz = dLz/dt |
You probably recall that potential energy is a useful concept, but that not every force has an associated potential energy. Forces which can be associated with a potential energy are called conservative forces. You probably recall that gravity (constant or 1/r²) and the electrostatic force are conservative, while friction isn't. We'd like a simple way of differentiating conservative forces from non-conservative forces.