PHY5210 W08

Chapter 10: Rotational Motion of Rigid Bodies

Reading:

In Taylor, read sections 10.1 for today, and 10.2 for Wednesday.

Recall

For a single particle, p = mv, L = r × p, and T = ½ m v².

Properties of the Center of Mass

In chapters 3 and 4 we derived "Newton's Laws" for extended objects, with particular emphasis on the laws for rigid bodies. We found that we can treat the motion of a rigid body in two parts:

the motion of the center of mass treated as a point particle with a mass equal to the total mass of the object (trajectory or orbit), and
the motion of the object about the center of mass (rotation).

But the derivation for the second part was basically limited to rotations about a fixed axis. We now wish to generalize this result to rotation about an arbitrary, and not necessarily fixed axis. Begin by recalling the definition of the center of mass and total momtentum of an object.

The center of mass is given by the vector expression

R_CM = (1/M) ∑_i m_i r_i,

and the total momentum of the object is

P = ∑_i p_i = ∑_i m_ir_idot = MR_CMdot,

where the sums are over all the individual point masses that make up the object. For a continuous object, the sums can be taken over to integrals. For example, the position of the center of mass becomes

R_CM = (1/M) ∫ r dm,

where the integral is over the entire mass distribution (volume).

The Total Angular Momentum

Since we will often deal with the CM, it is useful to express the angular momentum in terms of the CM of the system. Begin by defining the position of a point relative to the CM, r'_i = r_i - R_CM, and likewise the velocity in terms of the CM velocity and the velocity relative to the CM, v'_i = V_CM - v_i. The total angular momentum is the sum of the angular momentum over all particles:

L_tot = ∑_i (R_CM + r'_i) × m_i (V_CM + v'_i)

This can be expanded to give four terms:

L_tot = R_CM × P_CM + R_CM × ∑_i m_i v'_i + ∑_i m_i r'_i × V_CM + ∑_i m_i r'_i × v'_i

This can be simplified by noting that the second and third terms on the right are zero, and the last term can be expressed in terms of individual momenta to get

L_tot = R_CM × P_CM + ∑_i r'_i × p'_i.

To reach this solution I've used ∑_i m_i v'_i = 0 and ∑_i m_ir'_i = 0. Can you see why these are true? Hint, try calculating R_CM by expressing the coordinates in terms of the CM coordinates and the coordinate of the CM.

The above result says that the total angular momentum of a system of particles is composed of two parts: the angular momentum of the CM and the angular momentum about the CM. The second part is what we normally think of as the rotation or spinning of something. The first part we might consider as the angular momentum due to the orbit, or path of our system. This is quite a convenient occurance; the center of mass is a special point indeed.

L_tot = L(motion of CM) + L(rotation about CM) = L_orbit + L_rot.

where L_orbit = R_CM × P_CM and L_rot = ∑_i r'_i × p'_i.

This is beautifully illustrated by the motion of a planet. The planet (or its CM) orbits the sun, and therefore has orbital angular momentum. It can also spin on an axis through the CM, giving it rotational angular momentum. Note also, that the directions of the orbital and rotational angular momenta don't have to line up. For the planets in our solar system, they generally don't (the earth's axis of rotation is tilted by 23.5° to the direction of its orbital angular momentum).

Kinetic Energy of a System of Particles and a Rigid Body

For a system of n particles the kinetic energy is the sum of the individual kinetic energies:

T_tot = ∑_i T_i = ∑_i ½m_iv_i²

Now if we express v_i in terms of the CM velocity and the velocity relative to the CM, 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.

v_i² = (V_CM + v'_i)⋅(V_CM + v'_i) = V_CM² + v'_i² + 2V_CM⋅v'_i

Inserting this into the kinetic energy expression, and noting that the remaining dot product (third term) sums to zero, we get:

T_tot = ½MV_CM² + ∑_i ½m_iv'_i² = T_CM + T_rel

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.

T_tot = T_CM + T_rot

An alternative way to think of the kinetic energy comes from realizing that the derivation, up to the simplification of the expansion for v_i², doesn't depend on using the CM. That is, we can replace V_CM by the velocity of any point on the object, V, such that v_i = V + v'_i, and write the kinetic energy as

T = ½∑m_iV² + ½∑m_iv'_i² + V_CM⋅∑m_iv'_i.

Now if we choose the reference point to be one with zero (instantaneous) velocity such that V=0, the kinetic energy is just the second term

T = ½∑m_iv'_i².

Written this way, we find that the kinetic energy of the object is the rotational energy about any fixed point.

Potential Energy of a Rigid Body

We can write the potential energy for the conservative forces on a system. (Any non-conservative forces must be handled separately.) Write the potential as the sum of two parts, the potential energies due to internal and external forces separately, U = U^ext + U^int.

The internal potential energy depends on the relative separation of masses in the object, U^int = ∑_i<j U_ij(r_ij) where r_ij is the separation between two particles. In a rigid body, the separations are fixed, therefore the internal potential energy is a constant and can be neglected. The overall potential energy then reduces to U = U^ext.