Angular Momentum Incorrect Graph?

[gv3]

1. Homework Statement
Determine the total magnitude of angular momentum Ho of the particle about point O. The velocity of the particle is 5.5 m/s.

Homework Equations

Ho= r x mv

The Attempt at a Solution

The answer is 43.04. My question is, isn't the graph wrong? If you take the magnitude of the vector shown in the graph, it doesn't equal 5.5 it equals 6. Without the proper vector from the graph isn't the problem unsolvable? I only knew the answer because i was asked this question twice.

 

[gneill]

The only vector I can see indicated in the figure is V, which is shown to be directed along the path AB. AB itself is not the velocity vector, it just indicates its direction.

 

[gv3]

The only vector I can see indicated in the figure is V, which is shown to be directed along the path AB. AB itself is not the velocity vector, it just indicates its direction.

But isn't that vector 2i-4j-4k?

 

[gneill]

But isn't that vector 2i-4j-4k?

A position vector from A to B might be that, and it would have the same direction as V. But it is not V.

 

[gv3]

A position vector from A to B might be that, and it would have the same direction as V. But it is not V.

My mistake you are right. How would the velocity vector be found then? My professor in his lecture notes looks like he just took the direction of V shown.

 

[gneill]

All you need to take from the vector AB is its direction. Are you familiar with unit vectors? Can you form a unit vector that has the same direction as AB?

 

[gv3]

All you need to take from the vector AB is its direction. Are you familiar with unit vectors? Can you form a unit vector that has the same direction as AB?

yeah it would just be the position vector divided by the magnitude.