Recent content by myko

Ya. I see it. Thankyou. So then, would it be correct to proceed this way? Sincé the center of mass is moving with velocity v and there is no external forcé acting, it will keep moving same velocity. So it will travel distance $$d_{cm}=v(t1-t0)$$ from it's initial position. So at t1 it will be...

1. Two small particles of mass m1 and mass m2 attract each other with a force that varies with the inverse cube of their separation. At time t0 , m1 has velocity v directed towards m2 , which is at rest a distance d away. At time t1 , the particles collide. How far does m1 travel in the time...

I added some letters that where missed, sry. The momento of inertia of horizontal bar is I=md^2/12 and vertival bar I=md^2/3 so combined kinethik energy at position B is: K=((1/2)md^2/12+(1/2)md^2/3)\omega^2=2mg(d/2) is this correct?

I agree, that height lost is h=L/2, but the potential energy is 2mgh, because of the combined mass. But this is what I have written in the equation of my question. This leads me nowhere. I have 1 equation 2 unknowns..

picture The corresponding image for the situation

1. A T-shaped assembly is made of two identical uniform bars of length d and mass m each. One end of one of the bars is rigidly connected to the midpoint of the other. The assembly is pivoted at the connection point and held in position A. After it is released, the assembly swings down in the...

Ty Nathanael for your solution. I ,realy think, it is also a good and easy way to solve this. Actualy I think this is the way the problem was intended to be solved in the test I took. Thankyou again. What regard the asumption of constant acceleration, constant force implies constant...

Could you show how did you do it?

Ok. So here is the energy equation: $$1/2Mv^2+1/3Mv^2=Mgh$$ from which I get this relationship; $$h=\frac{5}{6g}v^2$$ And now, I supose, I have to express the distance traveled on the incline.$$d=h/\sin\theta$$ Now I could find the acceleration $$a=v^2/2d=v^2sin\theta/2h$$ and from here I can...

A hollow sphere with mass $$M$$ , radius $$R$$ , and moment of inertia $$I=2/3MR^2$$ about its center rolls without slipping with a initial center of mass speed $$v$$ towards a fixed ramp. It then rolls without slipping up the ramp. The ramp forms an angle $$\theta$$ with the horizontal...

Ok, ty. I got it.

$$mg(2r)=1/2mv^2+1/2I\omega^2+mgh$$$$h=\frac{2rg-7/10v^2}{g}$$ also $$h=1+cos\theta$ and now I would substitute v. Is that right?

[>A heavy sphere of radius r = 1.00 meter is fixed with respect to the ground. A small uniform solid sphere is placed at the top of the larger sphere. After a slight disturbance, the smaller sphere begins to roll downward without slipping. How high h is the small sphere above the ground at the...