Recent content by JuanMa2409

Maybe you're right, but I'd like to also prove it.

You're right, that's a easier way. So the acceleration in ##y## is: $$\ddot y=−(R\cos \theta)\dot \theta$$ And ##\dot \theta## is the angular speed ##\omega##, so: $$\ddot y=−(R\cos⁡ \theta)\omega$$ $$=−\frac{R\cos⁡ \theta)v^2}{R^2}$$ $$=−\frac{v^2\cos⁡ \theta}{R}$$ Knowing the velocity of the...

Thanks for your tip, I tried that method and I don't get any result yet. Here is my attempt: Where: - a_x is acceleration in x. - a_x is acceleration in y. - T is tension. I know that I need to calculate the points where the acceleration is purely horizontal, so I think that I'm searching for...

To solve this problem first I try to use polar coordinates, then I write the forces that I obtain in the free body diagram. That are the gravitational force and the tension force. With this using the second Newton's law I write the forces that are equal to the acceleration in polar coordinates...

Hello to everyone, I'm Juan Manuel. I'm a physics student, and I'm so glad of begin in this community.