Physics 53Rotational Motion 3Sir, I have found you an argument, but I am not obliged to find you an understanding.— Samuel JohnsonAngular momentumWith respect to rotational motion of a body, moment of inertia plays the same role that mass plays in the translational motion of a particle. It measures the intrinsic reluctance of a body to have its state of rotation changed.Torque plays the role in rotation that force plays in the translational motion of a particle. It describes the external influence that causes changes in the state of rotation.But what describes “the state of rotation” itself? For a particle, the state of translational motion is described by the linear momentum, p = mv.The corresponding quantity for rotational motion is the angular momentum.Like torque, angular momentum has meaning only with respect to some specified reference point. Also like torque, its magnitude depends on the distance from that point.We begin with the simplest system, a single particle. Later we will generalize to systems of particles, with special interest in rigid bodies. The definition for a particle is:Angular momentum of a particle L = r × pHere r is the position vector of the particle relative to the reference point, and p = mv is its linear momentum.Some properties of L:•L is a vector, perpendicular to the plane containing r and p, and thus perpendicular to both r and p.•L is zero if the particle moves along the line of r, i.e., directly toward or away from the reference point.PHY 53! 1! Rotations 3•The magnitude is given by L = r⊥p, where r⊥ is the moment arm, defined to be the perpendicular distance from the reference point to the line along which the particle moves.•Alternatively, L = rp⊥, where p⊥ is the component of p perpendicular to r.•L is a maximum, equal to rp, if p is perpendicular to r. This is the case if the particle moves (at least momentarily) in a circle about the reference point.Like linear momentum and kinetic energy, angular momentum is an important aspect of the state of motion of a particle, especially of orbital motion around some center of force. It is also an important property of the behavior of a system of particles.Torque as a vectorHere is the general definition of the torque of a force about a given reference point:Torque τ = r × FHere r specifies the location, relative to the reference point, of the point at which the force F is applied.Some properties of τ:•Torque is a vector, perpendicular to the plane containing r and F, and thus perpendicular to both r and F.•The torque is zero if F acts along the line of r, i.e., directly toward or away from the reference point.•The magnitude is given by the two formulas introduced earlier: τ= rF⊥= r⊥F.•The magnitude is a maximum, equal to rF, if F is perpendicular to r.The total torque on a system of particles is the sum of the torques of all forces that act on any of the particles in the system. We will see that the torques due to internal forces cancel, so the total torque is actually the sum of the torques due only to external forces.PHY 53! 2! Rotations 3Circular motion revisitedNow we apply these new definitions to a familiar problem, that of a particle constrained to move in a circle. Shown is a particle, attached to a massless rod of length r which is pivoted at the reference point, so that the particle moves in a circle about that point. We see that L = r × p has magnitude L = rmv and direction parallel to the angular velocity ω. Since v = rω, L = mr2ω= Iω. In vector form L = Iω. This simple relationship between L and ω also holds, as we will see below, in many — but not all — important situations involving systems of particles. It is the rotational counterpart of p = mv.Now consider the action of an external force applied to the particle. Let the force be tangent to the circle. (Other components would be counteracted by forces exerted by the rod.) If the force is parallel to the velocity, the particle speeds up; the torque τ = r × F is directed parallel to ω. Its magnitude is τ= rF. This force and the torque it produces give rise to a tangential acceleration ( F = mat) and to an angular acceleration (since at= α × r). The magnitudes obey F = mrα, so τ= mr2α= Iα. In vector form: τ = Iα. This simple relation between τ and α also holds in many — but not all — important situations involving systems of particles. It is the rotational counterpart of F = ma.These relatively simple properties of a single particle moving in a circle carry over to the rotational motion of a rigid body — provided it is a symmetric body rotating about its symmetry axis. This will be shown below.If L = Iω and τ = Iα (as in the case here) then because α = dω /dt (by definition) and I is a constant, we see that τ = dL/dt. As we will show below, this relation is true in all cases. It is the rotational counterpart of F = dp/dt.Angular momentum of a systemHere are some general properties of torques and angular momentum for any system of particles. The proofs of these statements are given at the end of this section.Total angular momentum Ltot= rCM× MvCM+ L(rel. to CM)ω r vPHY 53! 3! Rotations 3Angular momentum — like other properties of a system — breaks up into the sum of two terms: what the angular momentum would be if the system were a single mass point at the CM, plus the angular momentum as measured in the CM frame.Angular momentum (fixed axis) L( to axis) = IωWhat about the components of L perpendicular to the axis? In general they can behave in a quite complicated way, changing with time as the body rotates. These changes must be brought about by external torques, caused by forces exerted on the body by the fixed axle about which it rotates. But if the body is symmetric about the axis, these perpendicular components of L contributed by the “mirror image” mass points on opposite sides of the axis cancel each other exactly, and no external torques are required.For a symmetric body rotating about its symmetry axis, the angular momentum is entirely parallel to the axis and is equal to Iω.Most of the cases considered in this course involve such symmetric bodies. An example of a body that is not symmetric is an unbalanced wheel on a car. The friction caused by the normal forces exerted by the axle can rapidly wear out the wheel bearings. This is why one has the wheels “dynamically balanced” to make