Recent content by igor123d

Stupid Calculator! Lol, thanks a lot. I knew there must have been something really idiotic at the bottom, but being a bit OCD I couldn't rest until I found the cause, and I wasted more than an hour with this problem :). Well thanks again, now I can sleep.

I have some stupid trouble with a simple integration. f(x)=ln(x^(1/2))/x I try using u substitution. u=ln(x^(1/2)) Then du=1/(x^(1/2)*1/(2x^(1/2))dx=dx/(2x) Then dx should be 2xdu Then plugging back in I should have intg(2u*du) which would give me (ln(x^(1/2))^2; yet the answer my calculator...

Yes, I guess they will. Ok thanks for your help Dan, this makes sense now.

Well I guess since the string moves around the yo-yo, its distance would be R*theta. Then would it also have a translational part x, so that d=R*theta+x But since the rotation is not what's casuing the linear motion, how would x be related to r*theta?

But would this work with conservation of energy? I mean if the only force is F then the total work would have to be Fx. But in this case, you would also have rotational energy, so you would have Fx+K rotational. How could this work?

By the way here is an image of a similar situation (look at b): http://www.fas.harvard.edu/~scdiroff/lds/NewtonianMechanics/Yo-yo/Yo-yo002.gif

Well but in this specific example, the force actos horizontally on a point below the center of mass, so there would have to be torque, right?

Oh sorry forgot to mention, to the right.

1. A yo-yo lies on a frictionless table. If you apply a horizontal force F to the string to the right, how would the yo-yo move (linearly and rotationally) 2. EF=ma ET=I(alpha) 3. Initially I thought that the yo-yo would move to the right and rotate counter-clockwise (because the string lies...