Physics 1110 Fall 2004L40 - 1Lecture 40 29 November 2004Announcements• Reading Assignment for Wed, Dec. 1: Knight 14.6-14.8Vector Cross Product Just as in the case of calculating work, where we introduced a newvector product that was useful when we needed the parallelcomponents of vectors, in the case of angular dynamics we willintroduce a new vector product which is sensitive to perpendicularcomponents.Consider two vectors r A and r B , shown below, with an angle between them. The vector cross product r A r B is defined as a vector(not a number) with the magnitude of vector r A times the magnitude ofthe component of r B which is perpendicular to r A . The direction of thisnew vector is given by the right hand rule (RHR), that is, point yourfingers of your right hand in the direction of r A , curl them in thedirection of r B , and then your thumb points in the direction of thecross product vector. This cross product vector is perpendicular toboth r A and r B , and therefore is perpendicular to the plane that r A and r B define.Given these definitions, the magnitude of r A r B is AB sin which is infact equal to both AB and AB.Physics 1110 Fall 2004L40 - 2This new product has a couple of unusual features from regularmultiplication. First note that with this definition, any vector crossedwith itself (for example r A r A ) will give zero, since a vector cannothave a component perpendicular to itself. Second, note that the orderis important, that is, this operation is not commutative: r A r B r B r A .In fact, if you switch the order and try the right hand rule, you find thatthe sign changes. So the correct relation is r A r B = r B r A .Therefore, when you are calculating cross products you must keepthe correct order (or if you change the order, change the sign).Finally, if the two vectors are already perpendicular, then themagnitude of the cross product is r A and r B is just AB.If you have the magnitudes of the two vectors and the angle betweenthem already, then usually the easiest way to calculate the crossproduct is to use the formula with the sine above. However, as usual,to do precise work with arbitrary vectors you need to usecomponents. In this case, the cross product is a little tedious. I’ll showyou two ways to proceed.First take the case that r A and r B lie in the xy plane: r A = Axˆ i + Ayˆ j r B = Bxˆ i + Byˆ j We can calculate the cross product using the distributive principle: r A r B = (Axˆ i + Ayˆ j ) (Bxˆ i + Byˆ j )= AxBx(ˆ i ˆ i )+ AxBy(ˆ i ˆ j )+ AyBx(ˆ j ˆ i )+ AyBy(ˆ j ˆ j )= AxBy(ˆ i ˆ j )+ AyBx(ˆ j ˆ i )= (AxBy AyBx)(ˆ i ˆ j ).We used both principles about the cross product here. Finally, weactually define the cross product ˆ i ˆ j ˆ k ; this is the definition for aright-handed coordinate system, which we must use with the RHRrule to be consistent. So our final answer is that the cross product isPhysics 1110 Fall 2004L40 - 3in the z direction (perpendicular to the xy plane as it should) with thesize AxBy AyBx (which might be positive or negative).Since the cross product is intrinsically a 3 dimensional type ofoperation, we really need the full calculation for vectors with 3components. You can perform the same calculation above for thiscase, but there is a nice algorithm that is somewhat simpler or maybejust less tedious. Given the vectors r A and r B , form the following 3x3matrix:ˆ i ˆ j ˆ k AxAyAzBxByBzNow if you know something about matrix algebra, then if youcalculate the determinant of this matrix, it is in fact the cross product.If you don’t know about matrix algebra, continue on. The next step isto copy the first two columns to the right of the original matrix, makinga 3x5 matrix:ˆ i ˆ j ˆ k ˆ i ˆ j AxAyAzAxAyBxByBzBxByNow to find all the terms in the cross products you just draw adiagonal line down to the right starting from the top element of thefirst three columns, and then you repeat with a diagonal line up fromthe bottom element of the first three columns, but these three termsget a minus sign. Each term is made up of the product of the threeelements along the diagonal line. So for example, the first term isˆ i AyBz, the second term is ˆ j AzBx and the third term is ˆ k AxBy. Notethat these are all cyclic permutations of the sequence xyz incomponents. The terms from the upward diagonals are BxAyˆ k ,ByAzˆ i , and BzAxˆ j . When we put all of this together you get thefollowing (long) expression: r A r B = (AyBz AzBy)ˆ i + (AzBx AxBz)ˆ j + (AxBy AyBx)ˆ k .Physics 1110 Fall 2004L40 - 4Angular Momentum and Torque VectorsNow that we have the vector cross product, we can write thedefinitions of angular momentum and now torque quite elegantly: r L =r r r p r =r r r F The vector r r points from our axis of rotation to a particle withmomentum r p in the case of angular momentum. In the case oftorque, the vector r r points from the axis of rotation to the pointwhere the force r F is applied. Note that this does not change themagnitude of the torque we have been calculating but it does give it adirection in space, which is in fact parallel to the axis that the torqueis trying to cause rotation about. The directions of angular momentumand torque always come out along an axis, rather than in the plane ofthe motion, and this can seem very strange at first (or even second).Angular Momentum for a System of MassesNow that we have the angular momentum for a point mass relative tosome axis, we can calculate the total angular momentum for asystem of masses, which would in general include a solid extendedobject. If we start out with a number of point masses labeled i thenthe total angular momentum is the sum of the individual angularmomenta: r L total=r L ii= (r r iir p i).Since this is a vector sum with the cross product in it, it is not veryeasy to generalize it or go over to an integral. There is one simplecase however. If we have a solid object rotating about an axis ofsymmetry of the object, then the angular momentum r L is just Ir ,where I is the moment of inertia about that axis, and r is the angularvelocity (magnitude , direction given by RHR).Physics 1110 Fall 2004L40 - 5Conservation of Angular MomentumIn the absence of external torques, angular momentum is aconserved vector quantity. We saw this last week with the
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