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Spiral path in polar coordinates problem

  • Thread starter Koi9
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  • #1
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A bee goes from its hive in a spiral path given in plane polar coordinates by
r = b*ekt , θ = ct,
where b, k, c are positive constants. Show that the angle between the velocity vector and the
acceleration vector remains constant as the bee moves outward. (Hint: Find v · a/va.)


attempts.

v=r[dot]r[hat]+(r)θ[dot]θ[hat]

a=(r[double dot]-rθ[dot]^2)r[hat]+(rθ[double dot]+2r[dot]θ[dot])θ[hat]

r[prime]=bke^kt

r[double prime]= bk^(2)e^kt

θ[prime]=c

I know that v · a/va = vacosθ/va = cosθ, but I am unsure what to do with this knowledge or where to go from here,

thanks guys
 

Answers and Replies

  • #2
49
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Well, if the angle between the two vectors is a constant, then the time derivative of that angle should be constant, correct?

Also, be careful. Is it correct to say that the angle between the two vectors ([itex]cos\phi[/itex]) is the same as the angle given to you ( [itex]\theta = ct[/itex]).
 
Last edited:
  • #3
11
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Ah I see, so I should basically show that (va)cos(theta)/va cancels out to leave cos(theta), the time derivative of which is constant, showing that my angle is constant?
 
  • #4
11
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Also, if the angle between my two vectors is not theta, would it be v, since that seems to point in the same direction?

EDIT: So I want to show that cosθ=va/va, and that this equals a constant, so is a constant
 
Last edited:
  • #5
49
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Remember what a dot product is:
[itex] \vec{r_{1}}\cdot \vec{r_{2}} = x_{1}x_{2} + y_{1}y_{2} + z_{1}z_{2} = |r_{1}||r_{2}|cos \phi[/itex]
You can either sum of the product of the components of the two vectors or multiply there magnitudes times the angle between the vectors. You can use the above equation to solve for [itex]\phi[/itex] and see if it has any time dependence. I am not convinced that the angle between the velocity and acceleration vectors is [itex] \theta = ct [/itex]. And if it were, [itex]\theta[/itex] is not constant in time.
 
  • #6
49
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EDIT: So I want to show that cosθ=va/va, and that this equals a constant, so is a constant
Precisely.
 

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