# Understanding Angular Momentum#

There is a particle with mass $${{params_m}} \rm{kg}$$ hat is travelling with a speed of $${{params_v}} \rm{m/s}$$ in the direction $$\theta = {{params_theta}}^\circ$$ with the $$x$$-axis. It starts out at $$({{params_x}},0)$$.

## Part 1#

Calculate the magnitude of the angular momentum of this particle after $${{params_t}}$$ seconds.

Please enter in a numeric value in $$\rm{kg.m^2/s}$$.

You might have noticed that the angular momentum after $${{params_t}}$$ seconds is the same as the angular momentum at $$t = 0$$ seconds. This is because of the principle of Conservation of Momentum. There was no external moment affecting the particle hence we have Conservation of Momentum.
It can be seen if we actually go ahead and find the coordinates after $${{params_t}}$$ and calculate the angular momentum that way. $$x = x_0 + v.cos(\theta).t$$ where we find that $$x = {{params_x2}}$$ and similarly $$y = v\*sin(\theta)t$$ where $$y = {{params_y2}}$$ m. Now let us find the angular momentum with our new numbers. $$\vec{H} = x.\widehat{\mathbf{i}}\times m.v.sin(\theta)\widehat{\mathbf{j}} + y.\widehat{\mathbf{j}}\times m.v.cos(\theta)\widehat{\mathbf{i}}\Rightarrow \vec{H} = {{params_H1}}.\widehat{\mathbf{k}}-{{params_H2}}.\widehat{\mathbf{k}}$$ which is equal to $${{params_H0}}$$ $$\rm{kg.m^2/s}$$.
This is the same as if we did $$\vec{H} = {{params_x}}.\widehat{\mathbf{i}}\times m.v.sin(\theta)\widehat{\mathbf{j}}$$ where we also get $${{params_H0}}$$ $$\rm{kg.m^2/s}$$.