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.

Answer Section

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

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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 \({{params_H0}}\) \(\rm{kg.m^2/s}\).

This demonstrates that nature will always tend to conserve momentum as long as there are no external forces or moments.

Attribution

Problem is licensed under the CC-BY-NC-SA 4.0 license.