Can't work out integral in polar coordinates

In summary, the conversation discusses the work done by the frictional force in an infinitesimal angular displacement. The equation for this work is found to be dW = Frdθ = (krω)rdθ = kr^2(dθ/dt)dθ. The individual then tries to integrate this quantity but is unable to, and asks for tips. Later, they make some progress and get a new equation for v, but there is a sign error and a mistake in the exponent of r.
  • #1
etotheipi
Homework Statement
I made up a question where a car is moving around a quarter circle (from angle pi/2 to 0), where a tangential frictional force of magnitude kv acts in a direction opposite to the car's motion. It begins at the top of the quarter circle with speed u, and I want to work out its final speed. I chose to go about this in terms of energy considerations.
Relevant Equations
$$v = r \frac{d\theta}{dt}$$$$W = \int F dx$$
I considered the work done by the frictional force in an infinitesimal angular displacement:
$$dW = Frd\theta = (kr\omega) rd\theta = kr^{2} \frac{d\theta}{dt} d\theta$$I now tried to integrate this quantity from pi/2 to 0, however couldn't figure out how to do this$$W = kr^{2}\int_{\frac{\pi}{2}}^{0} \frac{d\theta}{dt} d\theta$$I was wondering if anyone could give me any tips! Thank you!Edit

I've made a little progress, I've changed the differential to$$d(\frac{1}{2}mv^{2}) = kr^{2}v d\theta$$so$$m dv = kr^{2} d\theta$$and consequently$$v = \frac{kr^{2}\theta}{m} + C$$Does this look right to anyone?
 
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  • #2
I think in your final result for ##v##, the square exponent in ##r## need not be there, should be just plain ##r## (this should propagate all the way back to dK=dW equation at the top of the edit.)
 
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  • #3
Delta2 said:
I think in your final result for ##v##, the square exponent in ##r## need not be there, should be just plain ##r## (this should propagate all the way back to dK=dW equation at the top of the edit.)

Ahh yes you're absolutely right. Thanks a bunch!
 
  • #4
I also think there is a sign error because the force is supposed to be frictional i.e it opposes particle speed, it should be ##F=-kv##.
 
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FAQ: Can't work out integral in polar coordinates

Why is it difficult to work out integrals in polar coordinates?

Integrals in polar coordinates involve a different coordinate system and require a different approach compared to integrals in Cartesian coordinates. This can make them more challenging to work out for those who are not familiar with polar coordinates.

Can I convert the integral from polar to Cartesian coordinates?

Yes, it is possible to convert an integral in polar coordinates to Cartesian coordinates. This can sometimes make the integration process simpler, but it depends on the specific integral being solved.

What are the key steps for solving an integral in polar coordinates?

The key steps for solving an integral in polar coordinates are: converting the integral to polar coordinates, simplifying the polar expression, and using appropriate techniques such as trigonometric identities to solve the integral.

Are there any special cases to consider when working out integrals in polar coordinates?

Yes, there are some special cases to consider when working out integrals in polar coordinates, such as dealing with singularities or when the integrand is not defined for certain values of theta.

Can software be used to help solve integrals in polar coordinates?

Yes, there are many software programs and online calculators available that can help with solving integrals in polar coordinates. However, it is important to have a basic understanding of the concepts and techniques involved in order to effectively use these tools.

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