Show that dA/dt is constant in time

[nns91]

Homework Statement

A particle is traveling with a constant velocity v along a line that is a distance b from the origin O. Let dA be the are swept out by the position vector from O to the particle in time dt. Show that dA/dt is constant in time and equal to (1/2)*L/m, where L is the angular momentum of the particle about the origin.

Homework Equations

L= rmv

The Attempt at a Solution

Any hint on how to do this problem ?

 

[Delphi51]

You'll need a nice diagram showing O, the distance b straight down from O to B. Off to the side a bit you need a bit of the path of the particle heading straight for B. Draw lines from O to each end of this bit of path, to form the triangle the question talks of. Label the area dA.

You must get an expression for the area of the triangle. You'll need the base and height, one of which is easy. For the other, think about how far the particle goes in time dt at its uniform speed. Just a bit of playing to get that dA/dt = L/(2m).

 

[Delphi51]

rho=1/3*pi*r^2

Trying to wrap my mind around this. Why one-third of the area of a circle?

 

[Doc Al]

Why one-third of the area of a circle?

It's just nonsense--ignore it.

 

[inf_sheet]

Trying to wrap my mind around this. Why one-third of the area of a circle?

What you're working on seems to be Kepler's second law. It's used in the derivation of the Volume of a sphere, since this has to be determined to yield mass.

It's not the only way to approach, as there's no absolute. It's just the way I did it.

You might find it easier this way, though, since the derivations will come in handy when you start playing around with gravity and the constant yield vectors.

 

[inf_sheet]

It's just nonsense--ignore it.

The BOMB!

 

[nns91]

Ok. Here is what I have been trying to do.

so L=mrv. Thus, r= L/(m*v)

The area of triangle is (1/2)*base*height. Height will be "b" so the base will just be r then ?

the I substitute as: (1/2)*b*(L/(m*v)) and get b*L/(2*m*v).

Am I going the right direction ?

Any further suggestion ??

 

[Doc Al]

Ok. Here is what I have been trying to do.

so L=mrv. Thus, r= L/(m*v)

In general, L = r X mv = rmv sinθ. But r sinθ = b in this case.

The area of triangle is (1/2)*base*height. Height will be "b" so the base will just be r then ?

Imagine the particle sweeping out a right triangle that has height b. The position vector forms the hypotenuse, but what about the other side? Express that other side as a function of time. Then find the area of the triangle as a function of time.

(Draw yourself a diagram.)