Question

What is the angular momentum LA if rA = 4, −6, 0 m and p = 11,

15, 0 kg · m/s? (Express your answer in vector form.)

Answer 1

The angular momentum LA is (-90, -24, 54) kg · m²/s.

Step-by-step explanation:

The angular momentum (LA) of a particle is given by the vector product of the position vector (rA) and the linear momentum (p). In this case, rA = (4, -6, 0) m and p = (11, 15, 0) kg · m/s. Performing the vector product, we get:

LA = rA x p

= (4i - 6j + 0k) x (11i + 15j + 0k)

= (4 * 0 - 0 * 15)i + (0 * 11 - 4 * 0)j + (4 * 15 - 11 * 6)k

= -90i - 24j + 54k

Therefore, the angular momentum LA is (-90, -24, 54) kg · m²/s.

Answer 2

The angular momentum LA is calculated using the cross-product of the position vector rA = (4, -6, 0) m and the linear momentum vector p = (11, 15, 0) kg · m/s to get LA = (126, 66, 0) kg · m²/s.

Step-by-step explanation:

To calculate the angular momentum LA of a particle, we use the cross-product of the position vector rA and the linear momentum vector p. Given rA = (4, -6, 0) m and p = (11, 15, 0) kg · m/s, we perform the cross-product:

LA = rA × p

LA = (4 × 15) - (-6 × 11), (6 × 11) - (0 × 15), (0 × 15) - (4 × 11)

LA = (60 + 66, 66 - 0, 0 - 0)

LA = (126, 66, 0) kg · m²/s