Recent content by Ailo

Hi! I would appreciate your thoughts on something. :smile: Let's say you have a ring with radius R rotating with angular velocity \omega about a vertical axis. A little bead is threaded onto the ring, and the friction between the bead and the ring is negligible. The bead follows the ring's...

Ohh! Now I get it. It's (theta)*R, right? *palmslap

My best guess is to make a triangle with sides R, R and ds. Will that work?

That's the problem. I've never done a problem like this before...

Hi! Thx for the answer, but the problem is how to set it up. I should get an expression, integrate it, and end up with the answer mg*R. I've got the force as a function of the angle, and I don't understand how to integrate it over a distance. Maybe I didn't explain the situation good enough...

Homework Statement A small mass m is pulled to the top of a frictionless halfcylinder (of radius R) by a cord that passes over the top of the cylinder. (a) If the mass moves at a constant speed, show that F=mg cos(\theta). The angle is between the horizontal and the radius drawn to the mass...

Higer potential energy-->more mass? If I go from the cellar to the loft, does my mass increase? If not, then where is the potential energy stored? Thanks for helping an unsure student! =)

What I'm saying here is that the ball hits the ground a horizontal distance 1,414*R units from where it started. This satisfies the inequality. But b asked how far from the base of the rock the ball lands. Now, I don't have english as my mother tongue, but I assume that the distance from the...

Now b is straightforward. When the ball hits the ground, y=0. R-\frac{1}{2}gt^2=0 \ \Rightarrow \ t=\sqrt{\frac{2R}{g}} The ball has then traveled the horizontal distance v_it=\sqrt{Rg} \cdot \sqrt{\frac{2R}{g}}=\sqrt{2}R. So the answer is (\sqrt{2}-1)R. :smile:

When t=0, v_i>\sqrt{Rg} ! I think I actually understood that. A million thanks Hootenanny (and Mentallic also)! =)

v_i^2t^2+(R-\frac{1}{2}gt^2)^2>R^2 After a lot of algebra, this reduces to v_i^2>Rg-\frac{g^2t^2}{4} So how do I find the t I'm after?

Yes, now I follow. =)

Yes. gx^2=2v_i^2y

I only get v_it>R \Rightarrow v_i>R/t. Even if i subsitute in what I know for t, R/t can't be the minimum velocity. Or did you mean something else?

x must be greater than R.