The position and linear momentum of a particle are r = (4i+5k) m and p=(8k^)kgm/s. What is its angular momentum when its position vector is r = (1.5i+2.2j)m?
The angular momentum of a particle is given by the cross product of its position vector and its momentum vector:
L = r x p
where x denotes the cross product. Since the position vector and the momentum vector are given, we can calculate the angular momentum as:
r = (4i + 5k) m
p = (8k^) kg m/s
r x p = det(i, j, k; 4, 0, 5; 0, 8, 0) = 32i - 20j - 32k
So the angular momentum of the particle is:
L = 32i - 20j - 32k
Now we need to find the angular momentum when the position vector is r = (1.5i + 2.2j) m. To do this, we calculate the cross product of r with p:
r = (1.5i + 2.2j) m
r x p = det(i, j, k; 1.5, 2.2, 0; 0, 0, 8) = 17.6i - 12j + 3.3k
So the angular momentum when the position vector is r = (1.5i + 2.2j) m is:
L = 17.6i - 12j + 3.3k