Recent content by welssen

Yes it makes sense: in polar coordinates [itex] \hat{\phi} [\itex] is a unit vector perpendicular to [itex]\hat{r}[\itex]. Here is a picture of it dimensionally: http://i.gyazo.com/6685ade5d3ab158b496077f7481551ea.png

Hi there, I've been trying to solve the following problem, which I found looks pretty basic, but actually got me really confused about the definition of angular momentum. Problem The trajectory of a point mass m is described by the following equations, in spherical coordinates: r(t) = r_0 +...

Thanks. Oh I see... I thought that, as ω is specified to be given in s^(-1), we were expected to convert it into rad/s (so multiplying it by 2π)... But OK I understand it now. So actually (a) E = 2*38.5 = 77 J. (b) P = 89.7 W.

Very nice, thanks! This was a mistake indeed, it's supposed to be in the nominator. So just a recap for (a): W = ∫τ0 sin(ωt)dθ = -τ0 cos(ωt) → from 0 to π gives → Supplied Energy = W = τ0 cos(ωt) We must consider that ω in the cos term has unit s^(-1) so we multiply it by 2π to have it in...

Thank you! So I apply the following integral from 0 to 2π: W=∫τ0 sin(ωt)dθ Then : dθ/2π = dt/T (T:=total time=2π/ω) ⇒ dθ=ω dt ⇒ W = ∫τ0 sin(ωt)/ω dt = -τ0 cos(ωt)/(ω^2) Evaluating the integral from 0 to 2π gives 0, again. Or should I integrate from 0 to π for the up-down movement? Thanks...

Homework Statement In bicycling, each foot pushes on the pedal for half a rotation of the pedal shaft; that foot then rests and the other foot takes over. During each half-cycle, the torque resulting from the force of the active foot is given approximately by τ = τ0 sin ωt, where τ0 is the...