Mechanics problems, about the use of polar coordinates

In summary: You'll have to figure out what the dot product of the unit vectors is. You can solve for the other quantities in a similar way.
  • #1
Xenonnn
4
0

Homework Statement


The problem and answers are given in full through the following image:
[URL]http://ekdhl.net/files/mechanics.JPG[/URL]

Homework Equations


Equations 2/13 and 2/14 are these:
[itex]\textbf{v} = r'e_{r} + r\theta 'e_{\theta}[/itex]
[itex]\textbf{a} = (r''-r\theta '^{2})e_{r} + (\theta ''r + 2r'\theta ')e_{\theta}[/itex]

The Attempt at a Solution


I could solve the introductory problems that had to do with polar coordinates easily, and I have worked with both spherical and polar coordinates in calculus classes, and haven't had a problem with it. But I can't figure out what the relationship is between [itex]\beta[/itex] and that the other metal bar, that is, how their movements correlate. The only thing I could come up with, and have tried, is to use the cosine theorem and the taking its derivative with regard to time. But this did not work, and dimensional analysis would also confirm that it did not yield the answer that I wanted.

So I'm stumped, and would appreciate your help!
 
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  • #2
It would help if you showed details of what you tried. Using the law of cosines is probably what you want to do. If the units didn't work out when you differentiated, that means you differentiated incorrectly.

You might also consider the vertical position of point A in terms of r and θ and in terms of AB and β to get another relation between the variables.
 
  • #3
Starting out with the laws of cosines I think seems very good, if I could make it work. This is what I tried (I have looked over it a couple of times for errors, but I can't find anything apparent.):

[itex]r = \sqrt{2*0.15^{2}-0.15^{2}*cos(\beta)}[/itex]
[itex]r' = \frac{0.15^{2}*sin(\beta)*\beta '}{2\sqrt{2*0.15^{2}-0.15^{2}*cos(\beta)}}[/itex]

Substituting in [itex]\beta = 60[/itex] and [itex]\beta ' = 0.6[/itex] I get r' = 0.3181..
 
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  • #4
You left a factor of two out of the law of cosines:
[tex]c^2 = a^2 + b^2 - 2ab\cos C[/tex]
where C is the angle opposite side c.

Another approach would be to use the fact that the triangle is an isosceles triangle to express θ in terms of β.
 
  • #5
vela said:
Another approach would be to use the fact that the triangle is an isosceles triangle to express θ in terms of β.

Thank you very much, using this I could find both [itex]\theta '[/itex] and [itex]\theta ''[/itex].

I now think I understand this problem:
1) Find the radius correlation using the law of cosines.
2) Find the angle correlation using the fact that the triangle is isosceles.

But I still can't nail the first part. This is the law of cosines:
[itex]c^2 = a^2 + b^2 - 2ab\cos C[/itex]

But in my situation, a = b = 0.15, so [itex]a^2+b^2 = 2*0.15^2[/itex], which is what I wrote. :/

I'm also a little worried that this is not the intended solution, since it does not require the use of the equations that they suggest!
 
  • #6
You're missing a factor of 2 in the cosine term.

Differentiating the law of cosines will give you the right answer. I was just pointing out there are multiple ways to break this problem down.
 
  • #7
vela said:
You're missing a factor of 2 in the cosine term.

Differentiating the law of cosines will give you the right answer. I was just pointing out there are multiple ways to break this problem down.

I see it now. Thank you very much help!
 
  • #8
Moderator's note: thread moved to Engineering, Comp Sci, & Technology. Please post questions from engineering courses in Engineering, Comp Sci, & Technology, not Introductory Physics.
 
  • #9
Xenonnn said:
I'm also a little worried that this is not the intended solution, since it does not require the use of the equations that they suggest!
I agree. Differentiating the law of cosines probably isn't how you're expected to solve the problem. What I think you're supposed to do is write down the velocity and acceleration of point A in terms of β and its derivatives and in terms of r, θ, and their derivatives. This will give you the relationships you need to solve for the derivatives of r and θ.

If you use point B as your origin, the velocity will be equal to
[tex]\vec{v}_\mathrm{A} = \dot{r}_B\hat{e}_{r_B} + r_B\dot{\beta}\hat{e}_\beta[/tex]
where rB is the distance from A to B. Since rB is constant, its derivative vanishes and you're left with
[tex]\vec{v}_\mathrm{A} = r_B\dot{\beta}\hat{e}_\beta[/tex]
With C as your origin, you have
[tex]\vec{v}_\mathrm{A} = \dot{r}\hat{e}_r + r\dot{\theta}\hat{e}_\theta[/tex]
To solve for [itex]\dot{r}[/itex], you would calculate the following:
[tex]\dot{r} = \hat{e}_r\cdot \vec{v}_\mathrm{A} = \hat{e}_r\cdot(r_B\dot{\beta}\hat{e}_\beta) = r_b\dot{\beta}(\hat{e}_r\cdot\hat{e}_\beta)[/tex]
You'll have to figure out what the dot product of the unit vectors is. You can solve for the other quantities in a similar way.
 

Related to Mechanics problems, about the use of polar coordinates

1. What are polar coordinates and how are they used in mechanics problems?

Polar coordinates are a type of coordinate system used to define the position of a point in two-dimensional space. They consist of a distance (r) and an angle (θ) from a reference point known as the origin. In mechanics problems, polar coordinates are often used to describe the motion of objects in circular or rotational motion.

2. How do you convert from Cartesian coordinates to polar coordinates?

To convert from Cartesian coordinates (x, y) to polar coordinates (r, θ), you can use the following equations: r = √(x² + y²) and θ = tan⁻¹(y/x). This will give you the distance and angle of the point from the origin.

3. Can polar coordinates be used to solve problems involving forces and motion?

Yes, polar coordinates can be used to solve problems involving forces and motion. In these types of problems, the forces and motion are often circular or rotational, making polar coordinates a useful tool for describing and analyzing them.

4. What are some advantages of using polar coordinates in mechanics problems?

One advantage of using polar coordinates in mechanics problems is that they simplify the equations and make them easier to solve. They are also useful for visualizing and understanding circular and rotational motion, as well as describing the motion of objects in polar coordinates.

5. Are there any limitations to using polar coordinates in mechanics problems?

While polar coordinates are useful for certain types of problems, they may not be the best choice for all mechanics problems. For example, they may not be as useful for problems involving linear motion or motion in three-dimensional space. It is important to consider the specific problem and determine if polar coordinates are the most appropriate coordinate system to use.

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