The angular momentum and the moment of inertia

The angular momentum , also called moment of momentum , is a vector quantity which in physics is generally indicated by the letter\ Vec {L}L, and accounts for the rotation in space of a massive body . Be careful not to confuse the angular momentum\ Vec {L}L with mechanical moment (or moment of forces) \ Vec {M}M or with the linear momentum (called momentum) \ Vec {p}p​.

Suppose we have a material point PPequipped with momentum \ Vec {p}p​, or equivalently mass mmand speed \ Vec {v}v (so that, in accordance with its definition, \ vec {p} = m \ vec {v}p​=mv), which rotates around a point in space OROR, called the pole of rotation. We call\ Vec {r}rthe radius-vector joining the poleOROR with the point PP. The angular momentum of the point is then definedPP around the pole OROR the vector \ Vec {L}Lgiven by the carrier product  between \ Vec {r}r is \ Vec {p}p​:\ vec {L} = \ vec {r} \ \ times \ \ vec {p}L=r × p​For the definition of carrier product, \ Vec {L}L It owns:

  • Direction perpendicular to the plane identified by\ Vec {r}r is \ Vec {p}p​;
  • Direction indicated by the “right hand rule”: if you plan to “screw” the vector with your right hand\ Vec {r}r on the carrier \ Vec {p}p​ with your fingers around your thumb, the verse of \ Vec {L}L is the one indicated by the thumb:
  • Module equal to the area of ​​the parallelogram having sides\ Vec {r}r is \ Vec {p}p​that is r \ m \ v \ \ sin (\ theta)r m v s i n ( θ ), being \ thetaθthe convex angle between\ Vec {r}r is \ Vec {p}p​.

The angular momentum is therefore maximum if \ Vec {r}r is \ Vec {p}p​ are perpendicular to each other (i.e. if \ theta = \ frac {\ pi} {2}θ=2π​): this happens for example when the point PPit moves with circular motion and the pointOROR is the center of the circumference, as the tangential speed \ Vec {v}v it is perpendicular to the radius \ Vec {r}r. The angular momentum is instead null when the directions of\ Vec {p}p​ and \ Vec {r}r are parallel (i.e. when \ theta = 0θ=0): for example, when PPmoves straight ( uniformly , uniformly accelerated , but not necessarily: it is sufficient that the trajectory is on a single straight line) and the poleOROR is located on the trajectory of PP.

Notice how in \ Vec {L}L a lot of information is concentrated:

  • \ Vec {L}L it is directly proportional to the distance rr from the pole : the further away from the center of rotation, the greater the angular momentum developed
  • \ Vec {L}L it is directly proportional to the speed vv of the material point : the faster the point moves, the greater its angular momentum will be
  • \ Vec {L}L it is directly proportional to the mass mm of the material point : a point of double mass will develop, at the same radius and equal speed, a double angular momentum.
  • The rotation circumference is located on the plane perpendicular to \ Vec {L}L and passing through the point PP

We also note that once identified \ Vec {L}Lthe poleOROR can be moved on the straight line having direction parallel to \ Vec {p}p​ and passing through OROR same : for the definition of a vector product, any components of the radius-vector\ Vec {r}r on this line do not contribute to the determination of \ Vec {L}L. In fact, parallelograms having the same height and the same base also have the same area: in our case, the base is made up of\ Vec {p}p​, while the height is the distance of the rella LL from the point PP.

However, a rotating material point also has an angular velocity : how do angular momentum and angular velocity relate? The answer is the moment of inertia .

Let’s think about a point PP that rotates circular motion around a pole OROR and is equipped with an angular momentum (compared to OROR) \ vec {L} = \ vec {r} \ times \ vec {p} = \ vec {r} \ times (m \ vec {v})L=r×p​=r×( mv). As said before on the position ofOROR, we can assume that \ Vec {r}r is perpendicular to \ Vec {p}p​. The motion ofPP it is circular, and therefore it is characterized by an angular velocity \ omega = \ frac {v} {r} \ Rightarrow v = \ omega \ rω=rv​⇒v=ω r, and the angular momentum is instead L = r \ m \ vL=r m v. We rewrite the angular momentum formula using the expression of angular velocity:L = r \ m (\ omega \ r) = mr ^ 2 \ \ omegaL=r ω )=m rωThe amountmr ^ 2m r2takes the name of Moment of Inertia , and is indicated by the letterTHETHE, from which equality is deducedL = I \ \ omegaL=THE ω

We have seen how a force imparting a rotation is responsible for a mechanical moment. There is a very deep link between mechanical moment and angular momentum : we are now going to illustrate it.

Be PP a material point subject to a moment of forces \ Vec {M}M: \ Vec {M}M calculated with respect to a rotation axis toto and therefore with respect to a pole ORORlocated on this axis; say\ Vec {r}r the radius-vector that connects OROR to PP, ed \ Vec {F}F the force applied in PP, is\ vec {M} = \ vec {r} \ times \ vec {F}M=r×FNow let’s use the fundamental equation of dynamics to write \ vec {F} = m \ vec {a}F=mto within the mechanical moment:\ vec {M} = \ vec {r} \ times (m \ vec {a})M=r×( mto)We recall the definition of acceleration: it is the variation of speed \ Delta \ vec {v}Δv, which occurred in a period of time \ Delta tΔ t: \ vec {a} = \ frac {\ Delta \ vec {v}} {\ Delta t}to=Δ tΔv​. The formula of the mechanical moment then becomes\ vec {M} = \ vec {r} \ times \ left (m \ frac {\ Delta \ vec {v}} {\ Delta t} \ right) = \ frac {\ vec {r} \ times \ left ( m \ Delta \ vec {v} \ right)} {\ Delta t}M=r×( mΔ tΔv​)=Δ tr×( Δv)​We recognize in this expression a variation of angular momentum: if the angular momentum is \ vec {L} = \ vec {r} \ times (m \ vec {v})L=r×( mv), its variation, taking place in the interval \ Delta tΔ t, is exactly \ vec {r} \ times (m \ Delta \ vec {v})r×( Δv)! We therefore conclude that\ vec {M} = \ frac {\ Delta \ vec {L}} {\ Delta t}M=Δ tΔL​This relationship is so important that it replaces the fundamental law of dynamics in rotating systems ! In fact, it is equivalent to it (that is, assuming\ vec {M} = \ frac {\ Delta \ vec {L}} {\ Delta t}M=Δ tΔL​ how true we can deduce the formula \ vec {F} = m \ vec {a}F=mto), but the quantities involved are more easily obtainable for rotating systems. In the animation below, a moment of forces (in blue) is applied to a rotating body to modify its angular momentum (in green):


A force is therefore responsible for the variation of the angular momentum. From this statement we deduce that, in the absence of forces or in equilibrium , that is, in accordance with the principle of inertia, when the resultant of all the forces and all the moments of the forces is zero, the angular momentum does not present variations : in in other words, the angular momentum remains . It would be more correct to say that it “tends” to be preserved: as soon as a force intervenes, which causes a mechanical moment, the angular momentum changes.

This result, apparently very theoretical and devoid of practical evidence, is instead for all to see: just look at a wheel of a bike or a motorcycle. From standstill, to speedv = 0v=0, there is no angular momentum, and the wheel (together with the bike, motorbike, driver and so on), if it is not kept standing with some other means (typically our legs), will fall to the ground. But as soon as it starts spinning, the wheel “stays” on its feet: the angular moment is responsible for this\ Vec {L}Lwith which it is equipped. The vector\ Vec {L}L tends to be preserved, and, together with it, the plane perpendicular to it, on which the motorcycle is “stuck”: so long as \ Vec {L}L is present, the bike needs nothing else to stand up.

Another very important consequence of the conservation of the angular momentum is the fact that the planets have plane orbits: this fact is enunciated by the laws of Kepler .


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