What are the Polar Coordinates for this Problem in Homework Statement?

In summary, the conversation discusses a question involving angular velocity and acceleration. The person asking for help provides diagrams and equations they have used to try and solve the question, but is struggling to find the correct answer. They eventually realize that their mistake was using degrees instead of radians in their calculations.
  • #1
steven10137
118
0

Homework Statement


http://img127.imageshack.us/img127/2695/coord2pq5gm6.jpg

Homework Equations


[tex]\underline v = \dot r\;\underline e _r + r\dot \theta \;\underline e _\theta[/tex]
[tex]\underline a = \left( {\ddot r - r\dot \theta ^2 } \right)\underline e _r + \left( {r\ddot \theta + 2\dot r\dot \theta } \right)\underline e _\theta[/tex]

The Attempt at a Solution


Hi guys, been stuck on this question for a while now.
Here are the diagrams I made:

http://img183.imageshack.us/img183/9205/coord2pq5diag1rx9.jpg

http://img359.imageshack.us/img359/3100/coord2pq5diag2xt7.jpg

So from the second diagram:
[tex]\begin{array}{l}
\cos 15 = \frac{{ - \dot r}}{v} \\
\therefore \dot r = - v\cos 15 = - 50\cos 15 = - 48.296\;ms^{ - 1} \\
\end{array}[/tex]
then
[tex]\begin{array}{l}
\sin 15 = \frac{{ - r\dot \theta }}{v} \\
\therefore \dot \theta = \frac{{ - v\sin 15}}{r} = \frac{{ - 50\sin 15}}{{770}} = - 0.017\;\deg s^{ - 1} \\
\end{array}[/tex]

We are told in the question that the acceleration of P is zero, hence the components of acceleration must be zero:
[tex]\begin{array}{l}
0 = r\ddot \theta + 2\dot r\dot \theta \\
\Rightarrow r\ddot \theta = - 2\dot r\dot \theta \\
\therefore \ddot \theta = \frac{{ - 2\dot r\dot \theta }}{r} = \frac{{ - 2 \times - 48.296 \times - 0.017}}{{770}} = - 0.002\deg s^{ - 2} \\
\end{array}[/tex]

This isn't the right answer.
I think it could be do with my angles used because I thought angular velocity was meant to be measured in radians. All help will be greatly appreciated.


Thanks in advance.
Steven.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
bump.
Does anybody have any suggestions?
 
  • #3
Oh my god, I answered my own question.

I changed all my angles to radians and put them into the the same equations, gave my answer in rad/s^2 and bingo!

[tex]{\dot r = - 50\cos \left( {\frac{\pi }{{12}}} \right) = - 48.296\:ms^{ - 1} }[/tex]

[tex]{\dot \theta = \frac{{ - 50\sin \left( {\frac{\pi }{{12}}} \right)}}{{770}} = - 0.017\:rad\;s^{ - 1} }[/tex]

[tex]{\ddot \theta = \frac{{ - 2\dot r\dot \theta }}{r} = \frac{{ - 2 \times - 48.296 \times - 0.017}}{{770}} = - 0.002\;rad\;s^{ - 2} }[/tex]

thanks anyways.
 

Related to What are the Polar Coordinates for this Problem in Homework Statement?

1. What are polar coordinates?

Polar coordinates are a system used to represent points in a two-dimensional space. They consist of a distance from the origin (r), and an angle from a reference line, typically the positive x-axis (θ).

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

To convert from polar coordinates (r, θ) to Cartesian coordinates (x, y), you can use the following formulas:

x = r * cos(θ)

y = r * sin(θ)

3. What is the purpose of using polar coordinates?

Polar coordinates are useful when dealing with circular or symmetrical shapes, as they make it easier to describe the location of points in relation to a center point and a reference angle. They are also commonly used in physics and engineering, particularly in problems involving circular motion or forces.

4. Can polar coordinates be negative?

Yes, both the distance (r) and angle (θ) in polar coordinates can be negative. A negative distance indicates a point in the opposite direction from the origin, while a negative angle represents a point in the opposite direction from the reference line.

5. How do you plot points using polar coordinates?

To plot a point using polar coordinates, first identify the distance (r) from the origin and the angle (θ) from the reference line. Then, starting at the origin, count out the distance along the reference line and rotate the angle in the direction specified. The point where the line intersects with the angle is the plotted point.

Similar threads

  • Calculus and Beyond Homework Help
Replies
12
Views
1K
Replies
8
Views
266
  • Calculus and Beyond Homework Help
Replies
1
Views
846
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
Replies
8
Views
997
  • Aerospace Engineering
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Advanced Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
284
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
Back
Top