# Chapter 12 - Rotation of a Rigid Body - Exercises and Problems: 43

The magnitude of the angular momentum relative to the origin is $1.2~kg~m^2/s$ in the counterclockwise direction.

#### Work Step by Step

We can find the momentum vector of the particle. $p = (m~v)~\hat{j}$ $p = (0.10~kg)(4.0~m/s)~\hat{j}$ $p = (0.40~kg~m/s)~\hat{j}$ We can express the particle's position as a vector. $r = (3.0~m)~\hat{i}+(2.0~m/s)~\hat{j}$ We can find the angular momentum relative to the origin. $L = r\times p$ $L = [(3.0~m)~\hat{i}+(2.0~m/s)~\hat{j}]\times [(0.40~kg~m/s)~\hat{j}]$ $L = (1.2~kg~m^2/s)~\hat{k}+0$ $L = (1.2~kg~m^2/s)~\hat{k}$ The magnitude of the angular momentum relative to the origin is $1.2~kg~m^2/s$ in the counterclockwise direction.

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