Question

I have a question on Challenge Problem 3.3 on Module 3. I would greatly appreciate it if someone could show all the steps from the equation L_0=r_0,m x mv to the last equation, -6kg m^2 s^-1 k. I couldn't follow the steps from the solution, because they were too brief. Here's the link to the solutions: http://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/cartesian-coordinates-and-vectors/MIT8_01SC_problems03_soln.pdf. Thanks in advance!

Answer 1

see, we know
r = 2 i + 3 j........(i)
v = 3 i + 3 j
now mv = 2( 3i + 3j) = 6i + 6j................(ii)
now,
L = r x mv
= (2i + 3j) x (6i + 6j)
= (2i x 6i) + (2i x 6j) + (3j x 6i) + (3j x 6j)
= 0 + 12 k - 18 k + 0.........[ i x i = j x j = 0] [ 2i x 6j = 12sin90 k = 12 k]
= - 6 k
this is what happened there.
u got it?

Answer 2

I understand it now and see how to get the -6k. However, I don't get the part where you wrote: [ 2i x 6j = 12sin90 k = 12 k]. I just saw in the notes that i x j=k. What is the sin90 for? Thanks!

Answer 3

u know how |a x b| = |a||b|sin A
where A is the acute angle between vectors a and b. in this case, i and j are perpendicular to eachother. so sin90 = 1.
hence |a x b| = |a| |b|..........this is how we got 12. now k specifies the direction of the resultant vector. since the resultant is perpendicular to both the multiplying vectors, we get
i x j = k and j x i = -k.

Answer 4

Ok, but why do we take the magnitude of the cross product?

Answer 5

how else would u get the 12 in [2i x 6j = 12k] ?
here the vector 12k is made up of two parts - the magnitude = 12
- the direction = +k
u get the direction from right hand rule. to find the magnitude there are two ways:
1) from determinant
2) using the formula i used above.

Answer 6

Ok - thanks!