Question

At a particular instant, a 1.0-kg particle’s position is →r=(2.0ˆi−4.0ˆj+6.0ˆk)m, its velocity is →v=(−1.0ˆi+4.0ˆj+1.0ˆk)m/s, and the force on it is →F=(10.0ˆi+15.0ˆj)N.

What is the angular momentum of the particle about the origin?

a) −11kg⋅m^2 /s
b) −13kg⋅m^2/s
c) −15kg⋅m^2 /s
d) −17kg⋅m^2/s

Answer 1

The angular momentum of the particle about the origin is -11.0 kg⋅m²/s.

Step-by-step explanation:

The angular momentum of a particle about the origin can be calculated using the formula Ỉ = ŕ × p, where Ỉ is the angular momentum, ŕ is the position vector from the origin to the particle, and p is the linear momentum of the particle. In this case, the position vector is →r=(2.0ˆi−4.0ˆj+6.0ˆk)m and the linear momentum is →p=(−1.0ˆi+4.0ˆj+1.0ˆk)m/s. Taking the cross product of these vectors, we get:

Ỉ = (2.0ˆi−4.0ˆj+6.0ˆk) × (−1.0ˆi+4.0ˆj+1.0ˆk)

Simplifying the cross product, we get: Ỉ = −11.0ˆi + 7.0ˆj + 16.0ˆk kg⋅m²/s. Therefore, the angular momentum of the particle about the origin is −11.0 kg⋅m²/s.