Is the Angle Between Velocity and Acceleration Constant for a Spiraling Bee?

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Homework Help Overview

The problem involves a bee moving in a spiral path described in plane polar coordinates, with the goal of determining whether the angle between its velocity and acceleration vectors remains constant as it moves outward.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of angles between vectors and the relevance of using the dot product. There are attempts to clarify the expressions for velocity and acceleration in polar coordinates, with some questioning the completeness of the original poster's approach.

Discussion Status

Some participants have provided guidance on using the dot product to explore the constancy of the angle, while others express concerns about the complexity of converting between coordinate systems. Multiple interpretations of the problem and approaches to the solution are being explored.

Contextual Notes

There are mentions of guidelines for problem-solving and the need for clarity in mathematical expressions. Participants also note the potential challenges of transitioning between polar and Cartesian coordinates.

negation
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Homework Statement




A bee goes out from its hive in a spiral path given in plane polar coordinates by r = bekt q = ct where b, k, and c are positive constants. Show that the angle between the velocity vector and the acceleration vector remains constant as the bee moves outward.
Solution
r(t) = bekt q(t) = ct

The Attempt at a Solution



r'(t) = bke^(kt)
r' ' (t) = bk^(2)e^(kt)

I'm lost as to what should I be doing with the information to make the jump to the proof.
 
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Not good to throw up your hands under 3). See guidelines.

How does one go about calculating an angle between vectors ?
 
He, hello Negation!

Also some more explanation under 1) is needed. You should know better by now! I can deduce some things, so I can guess that ## r = b \, e^{kt} ## and ##\theta = q\, t## but this way you make it too difficult for others. At least learn a bit about Go Advanced. It is a safe, easy environment. ##\TeX## is much more powerful, but also a lot tougher. Still worth the investment...
 
Oh, and: if a vector in plane polar coordinates has two components, how come you only write down ##\dot r## and ##\ddot r## (this is shorthand for ##dr\over dt## and ##d^2 r \over dt^2##) if you really need ##\dot {\vec r}## and ##\ddot {\vec r}## to say something about the angle ?
 
BvU said:
Not good to throw up your hands under 3). See guidelines.

How does one go about calculating an angle between vectors ?

By using the dot product and which by definition states:

A.B = |A||B|cosΘ
Θ=arc cos[(A.B)/(|A||B|)]

I understand that there are 2 components. I trying to outline the idea first.
Suppose, if I can show that theta does not change, would I then have proven?
 
Right! the dot product divided by the magnitudes should give you something constant. From now on, it is math all the way to the answer.
 
Yes. Showing (A.B)/(|A||B|) is constant is sufficient.
Now, the exercise is written out in polar coordinates. You can do a lot of work transforming to cartesian, or you can stay with polar (r, Θ).
 
BvU said:
Yes. Showing (A.B)/(|A||B|) is constant is sufficient.
Now, the exercise is written out in polar coordinates. You can do a lot of work transforming to cartesian, or you can stay with polar (r, Θ).

I'm a little disturbed by the amount of hassle having to make the conversion from polar to cartesian and vice versa. It's tedious and messy. Do you recommend learning lagrangian in first year physics or should I really focus on the existing math and physics in school first?
 
Focus. But ##x=r \cos\theta## and ## y = r \sin\theta## is first year. So is the dot product in polar coordinates ## (r_1, \theta_1)\cdot (r_2, \theta_2) = |r_1|\, |r_2|\, \cos(\theta_1 - \theta_2)##

There is no way to avoid math in phys, I would say. Philosophers might dream otherwise.
 

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