- #1
StephenPrivitera
- 363
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The "x" means cross product
v=w x r
This is the tangential velocity.
So, a=(dw/dt) x r+w x (dr/dt)
(dw/dt) x r is the tangential acceleration
w x (dr/dt) is the radial (centripetal acceleration)
I learned this from class.
But where is the radial velocity?
If we write s=T x r, then
v=w x r+T x (dr/dt)
Can T x (dr/dt) be interpreted as radial velocity?
Now a=(dw/dt) x r + w x (dr/dt) + w x (dr/dt) + T x (d2r/dt2)
=(dw/dt) x r + 2w x (dr/dt) + T x (d2r/dt2)
Through a little geometry, I can show that dr/dt=w x r
So, a=(dw/dt) x r + 2w x (w x r) + T x (d(w x r/dt)
=(dw/dt) x r + 2w x (w x r) + T x ((dw/dt) x r + w x (w x r))
First question: have I violated physics and mathematics at once?
Second question: can I simplify this? (I don't know the distributive/associative properties of "x")
Third question: What do each of these terms signify? (ie, cent. accel, tang. accel, etc)
Fourth question: What exactly is this vector T?
v=w x r
This is the tangential velocity.
So, a=(dw/dt) x r+w x (dr/dt)
(dw/dt) x r is the tangential acceleration
w x (dr/dt) is the radial (centripetal acceleration)
I learned this from class.
But where is the radial velocity?
If we write s=T x r, then
v=w x r+T x (dr/dt)
Can T x (dr/dt) be interpreted as radial velocity?
Now a=(dw/dt) x r + w x (dr/dt) + w x (dr/dt) + T x (d2r/dt2)
=(dw/dt) x r + 2w x (dr/dt) + T x (d2r/dt2)
Through a little geometry, I can show that dr/dt=w x r
So, a=(dw/dt) x r + 2w x (w x r) + T x (d(w x r/dt)
=(dw/dt) x r + 2w x (w x r) + T x ((dw/dt) x r + w x (w x r))
First question: have I violated physics and mathematics at once?
Second question: can I simplify this? (I don't know the distributive/associative properties of "x")
Third question: What do each of these terms signify? (ie, cent. accel, tang. accel, etc)
Fourth question: What exactly is this vector T?
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