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# At time $t = 0$, a 3.0 kilogram particle with velocity ${\hat{v}} = (5.0\ \text{~m/s}){\hat{i}}\ -\ (6.0\ \text{~m/s}){\hat{j}}$ is at $x = 3.0\text{~m}, y = 8.0\text{~m}$. It is pulled by a $7.0 \text{~N}$ force in the negative x direction. About the origin, what is the particle’s angular momentum?

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Given:

The particle is 3 kg, moves with velocity $\vec{v} = (5\hat{i} - 6 \hat{j})\mathrm{ (m/s) }$ at $x = 3$ m and $y= 8$ m. The force is 7 N in the negative x direction.

$(a)~~$ The angular momentum with respect to the origin is defined as

\begin{aligned} \vec{l} = m(\vec{r}\times\vec{v}) \end{aligned}

Substitute the givens

\begin{aligned} \vec{l} & = 3 \mathrm{ (kg) } \times [( 3\hat{i} + 8 \hat{j})\mathrm{ (m) } \times (5\hat{i} - 6 \hat{j})\mathrm{ (m/s) }]\\ & = 3 \mathrm{ (kg) } \times [- 18 \hat{i}.\hat{j} + 40 \hat{j} \cdot \hat{i}] \mathrm{ kg \cdot m^2/s }\\ & =(-174\hat{k})\mathrm{ kg \cdot m^2/s } \end{aligned}

\begin{aligned} \fbox{ \vec{l} = (-174\hat{k})\mathrm{ kg \cdot m^2/s } } \end{aligned}

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