A 2.51 kg particle moves in the xy plane with a velocity of v = ( 4.98 , -3.58 ) m/s. Determine the angular momentum of the particle about the origin when its position vector is r = ( 1.73 , 1.15 ) m.

To determine the angular momentum of a particle about the origin, we need to calculate the cross product of its position vector and its linear momentum. The angular momentum formula is given by:

L = r x p

where L is the angular momentum, r is the position vector, and p is the linear momentum.

Given:
Mass of the particle (m) = 2.51 kg
Velocity (v) = (4.98, -3.58) m/s
Position vector (r) = (1.73, 1.15) m

First, we need to find the linear momentum (p) of the particle. The linear momentum (p) is given by:

p = m * v

where p is the linear momentum, m is the mass, and v is the velocity.

Calculating the linear momentum (p):
p = 2.51 kg * (4.98, -3.58) m/s
p = (12.5498, -9.0058) kg∙m/s

Now, we can calculate the angular momentum (L) using the cross product of the position vector (r) and the linear momentum (p):

L = r x p

To find the cross product, we can use the determinant method or the dot product method. Let's use the determinant method:

L = | i j k |
| r1 r2 r3 |
| p1 p2 p3 |

Since we are working in the xy plane, the z-component is zero, so we can ignore it.

L = (r1 * p3 - r3 * p2)i - (r1 * p3 - r3 * p1)j + (r1 * p2 - r2 * p1)k

Substituting the values:
L = (1.73 * 0 - 0 * -9.0058)i - (1.73 * 0 - 0 * 12.5498)j + (1.73 * -9.0058 - 1.15 * 12.5498)k

Simplifying:
L = -19.528k

Therefore, the angular momentum of the particle about the origin is -19.528 units in the k-direction.