- #1
Benny
- 584
- 0
Hi, can someone help me do the following question? (I've cut out some details, leaving the results which might be of help)
Let the vector r represent the displacement from the origin to a moving particle of mass m which is subjected to a force F.
Results which I've been able to arrive at:
[tex]
\mathop H\limits^ \to = \mathop r\limits^ \to \times \mathop p\limits^ \to \Rightarrow \frac{{d\mathop H\limits^ \to }}{{dt}} = m\frac{d}{{dt}}\left( {\mathop r\limits^ \to \times \mathop v\limits^ \to } \right) = \mathop r\limits^ \to \times \mathop F\limits^ \to = \mathop M\limits^ \to
[/tex]
v is velocity, F is force, p = mv is moment.
If F acts towards the origin then [itex]\frac{d}{{dt}}\left( {\mathop r\limits^ \to \times \mathop v\limits^ \to } \right) = \mathop 0\limits^ \to [/itex].
Now here is what I am having trouble with.
Show that [itex]\left\| {\mathop r\limits^ \to \times \mathop v\limits^ \to } \right\|[/itex] is twice the rate at which the area A is swept out by the vector r, that is,
[tex]
\left\| {\mathop r\limits^ \to \times \mathop v\limits^ \to } \right\| = 2\frac{{dA}}{{dt}} = r^2 \frac{{d\theta }}{{dt}}
[/tex]
and hence deduce that, when a mass is subject to a central force directed towards the origin, the vector displacement r moves in a plane and sweeps out equal areas in equal times.
I don't know where to start. I can't visualise what is going on in this problem. Norm of r cross v is the area of a parallelogram...ok...but that doesn't tell me much. As for r, it is arbitrary so I can't try to make a picture. I just don't know what to do here. Can someone help me?
Let the vector r represent the displacement from the origin to a moving particle of mass m which is subjected to a force F.
Results which I've been able to arrive at:
[tex]
\mathop H\limits^ \to = \mathop r\limits^ \to \times \mathop p\limits^ \to \Rightarrow \frac{{d\mathop H\limits^ \to }}{{dt}} = m\frac{d}{{dt}}\left( {\mathop r\limits^ \to \times \mathop v\limits^ \to } \right) = \mathop r\limits^ \to \times \mathop F\limits^ \to = \mathop M\limits^ \to
[/tex]
v is velocity, F is force, p = mv is moment.
If F acts towards the origin then [itex]\frac{d}{{dt}}\left( {\mathop r\limits^ \to \times \mathop v\limits^ \to } \right) = \mathop 0\limits^ \to [/itex].
Now here is what I am having trouble with.
Show that [itex]\left\| {\mathop r\limits^ \to \times \mathop v\limits^ \to } \right\|[/itex] is twice the rate at which the area A is swept out by the vector r, that is,
[tex]
\left\| {\mathop r\limits^ \to \times \mathop v\limits^ \to } \right\| = 2\frac{{dA}}{{dt}} = r^2 \frac{{d\theta }}{{dt}}
[/tex]
and hence deduce that, when a mass is subject to a central force directed towards the origin, the vector displacement r moves in a plane and sweeps out equal areas in equal times.
I don't know where to start. I can't visualise what is going on in this problem. Norm of r cross v is the area of a parallelogram...ok...but that doesn't tell me much. As for r, it is arbitrary so I can't try to make a picture. I just don't know what to do here. Can someone help me?