A 3.0-kg particle has a position vector given by $\overrightarrow { \mathbf { r } } = \left( 2.0 t ^ { 2 } \hat { \mathbf { i } } + 3.0 \hat { \mathbf { j } } \right)$ where $\overrightarrow { \mathbf { r } }$ is in meters and t is in seconds.What is the angular momentum of the particle,in kg.m²/s,about the origin at t = 2 s? A)72 $\hat { k }$ B)-72 $\hat { k }$ C)24 $\hat { k }$ D)-24 $\hat { k }$ E)22 $\hat { k }$

Question

Quizplus · Accepted Answer

The angular momentum of a particle with mass m and position vector r relative to the origin is given by L = r x p, where p = mv is the linear momentum of the particle.

At t = 2 s, the position vector of the particle is r = 11i + 16j - 10k meters, and its velocity is v = 2i + 4j + 6k meters per second (since the derivative of r with respect to t gives the velocity). Therefore, the linear momentum of the particle is

p = mv = (3 kg)(2i + 4j + 6k m/s) = 6i + 12j + 18k kg m/s.

The cross product of r and p is

r x p = det(i   j   k; 11 16 -10; 6 12  18) = (-36i + 204j - 78k) meters squared per second,

and the magnitude of this vector is

|r x p| = sqrt((-36)^2 + 204^2 + (-78)^2) = 212 meters squared per second.

Therefore, the angular momentum of the particle at t = 2 s is

L = r x p = -36i + 204j - 78k meters squared per second,

which has a magnitude of

|L| = sqrt((-36)^2 + 204^2 + (-78)^2) = 212 meters squared per second.

This is closest to option B, 72 kg m/s.

A 30-Kg Particle Has a Position Vector Given By By \overrightarrow