Recent content by lgmulti

I made a mistake. It should be $$T \cos \theta = F_c = \frac{mv^2}{l\cos\theta}$$

Oh, so the force ##T## must also satisfy both $$T \cos \theta = F_c = \frac{mv^2}{l\sin\theta}$$ and $$T \sin \theta = mg.$$ At this time if I want to express ##v## in terms of other variables, I can rewrite the first equation into $$Tl\cos\theta\sin\theta = mv^2$$ and plug the second equation...

So the object always need an upward force to maintain the circular motion, and it's usually not drawn on the FBD. In my case it's the tension and the gravity together that creates centripetal force. I got it. Thanks for everyone

I don't know how ##\theta## affects the speed of the mass. Is it infinity?

Oh, so is ##F_\rm{combined}## in my figure actually the centripetal force?

Oh, so the object can keep the uniform circular motion as long as $$F_c \sin \theta = mg.$$ But it seems impossible to happen when ##\theta = 0## because ##\sin 0 = 0##. Is it right?

So does it look like this when the point at the center of the circle is above the plane? If so, what will happen if the point at the center of the circle is on the plane

Here is the figure for my question. The object is doing uniform circular motion around the gray circle that is in a plane parallel to the ground, the pink force F_combined is the combination of the two red force G (gravity) and F_c (centripetal force), so the pink force will always point...

I can't upload image, so the problem is the same as the summary: When an object is doing uniform circular motion in a plane parallel to the ground, the centripetal force is parallel to the ground while the gravity is vertical to the ground, so the combination of the two force must be downwards...