Recent content by UnterKo

BvU, I really don't know what is it... could you please answer to my questions in 19? I am trying to figure out this problem for more than a week so that's the reason why I've asked you for help.

I've done it and I see what you mean. But there can not be found any max. lifting force at some angle? Eg. ##\theta = \frac{\pi}{2} ##(as measured from horizontal in downwards direction)?. So how can I find the value of ##X = \frac{m}{M}##? And could you, please, explicitly write which ones of...

It is maximised for ##\theta = \frac{\pi}{2}##. Is it centripetal force? So is it going to be part b): ##PE = mgR \sin \theta##?

I see, I have not understood this till now! :) So it's going to be a centripetal force? And the angle may be found by some derivation? Is the loss of PE going to be like: ##PE = mgh##, so for one of the beads at the bottom: ##PE = mg(\sin \frac{\pi}{2})##, so the loss is: ##PE = mg(2R - 1)##?

But then they are at the bottom of the hoop, no? And for this position the fraction m/M has to be determined, right?

So the force experienced by beads on the hoop is a sum of this force (N) of both beads? Then: 2mg(3sin θ - 2) > Mg. And then it is: m/M > (3 sin θ - 2)/(2)? So: K = ½ mR θ'2. But if I substitute, for example, m = M (3 sin θ - 2)/(2), I'm sorry, but I don't see how it's going to help me.

$$ N = mg (3 \sin \theta - 2)$$<- it isn't this multiplied by two? I'm not actually sure about what even physically means 'to rise up off the support'. Does it mean that the hoop is not fixed? b) KE before the collision is: $$ K = 1/2 ma^2 \dot{\theta}^2$$

a) Yes, it's upwards to the horizontal. When I substitute, it gets: $$ ½ ma^2 θ'^2 + mga \sin θ = mga$$ $$ ½ a θ'^2 + g \sin θ = g $$ $$ ½ a θ'^2 = g(1 - \sin θ)$$ $$ θ'^2 = \frac{2g(1 - \sin θ)}{a}$$ and in radial eg.: $$N - mg \sin θ = -2 mg (1 - \sin θ)$$ $$N = mg(3 \sin θ - 2)$$ But how do...

a) Each bead experiences its weight and the normal reaction force (because the hoop is frictionless, it acts normal to the surface, right?). If I take θ as the angle of it's position with respect to the horizontal, the normal force may be expressed as: F = Nr - mgk = (N - mg sin θ)r - (mg cos θ)...

Hello, I've got a problem and I have no idea how to start. I'll be happy for any hint. Thanks Homework Statement Two beads each of mass m are at the top (Z) of a frictionless hoop of mass M and radius R which lies in the vertical plane. The hoop is supported by a frictionless vertical support...