- #1
String theory guy
- 26
- 4
- Homework Statement
- Is there a way of solving this problem without without using polar coordinate's.
- Relevant Equations
- CM formula
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None of the replies mention polar coordinates.String theory guy said:Sorry I was not clear. I would like to understand how to solve this problem without using polar coordinate's.
Well, it is at least a symmetry argument that people have mentioned. I see three approaches via symmetry. There may be more.kuruman said:This is the symmetry argument that people have mentioned.
Yep, Thank You.kuruman said:Start with the definition of the x-position of the CM relative to the center, $$X_{\text{cm}}=\frac{1}{M}\int x~dm=\frac{1}{M}\int x~\lambda ds$$ where ##ds## is an arc element ##ds=\sqrt{dx^2+dy^2}.##
Now $$x^2+y^2=R^2\implies xdx+ydy=0\implies dy=-\frac{xdx}{y}.$$Therefore $$ds=\sqrt{dx^2+\frac{x^2 dx^2}{y^2}}=dx\sqrt{\frac{x^2+y^2}{y^2}}=R\frac{dx}{|y|}.$$Thus if we integrate over a semicircle from the 3 o'clock position to the 9 o'clock position, $$X_{\text{cm}}=\frac{\lambda R}{M}\int_{R}^{-R}\frac{x dx}{|y|}=\frac{\lambda R}{M}\int_{R}^{-R}\frac{x dx}{\sqrt{R^2-x^2}}.$$At this point you can
Obviously, the integral that completes the circle from the 9 o'clock position to the 3 o'clock position also vanishes and so do the integrals for ##Y_{\text{cm}}##.
- Observe that you have an odd function integrated over symmetric limits which means that it vanishes. This is the symmetry argument that people have mentioned.
- Actually do the integral and verify that it vanishes. Let ##u^2=R^2-x^2##, etc. etc.
Is this what you were looking for?
CoM stands for "Center of Mass", which is the point at which an object's mass is evenly distributed in all directions. In the case of a uniform thin hoop, the CoM is the point at the center of the hoop where the mass is evenly distributed around the axis of rotation.
The CoM of a uniform thin hoop can be calculated by finding the average of the radii of the hoop. This is known as the "radius of gyration". The CoM is located at a distance of half the radius of gyration from the axis of rotation.
No, the CoM of a uniform thin hoop does not change when the hoop is rotated. This is because the mass is evenly distributed around the axis of rotation, so the CoM remains at the same point.
The CoM of a uniform thin hoop is an important factor in determining its stability. If the CoM is located above the point of contact with the ground, the hoop will be stable and will not tip over. However, if the CoM is located outside the point of contact, the hoop will be unstable and will tip over.
Yes, the CoM of a uniform thin hoop can be outside the physical boundaries of the hoop. This is because the CoM is a theoretical point that represents the distribution of mass in an object, and does not necessarily have to be located within the physical boundaries of the object.