Recent content by chod

I think it means that the moon will be full as they are separated by 180 degrees (if the nodes were aligned then it would be a lunar eclipse). Thanks for the thread, I'll check it out

I feel as though I have a good understanding of the movements of the moon, sun, planets and any other bodies out there, but there is still one thing that has been giving me trouble. My problem is finding out the phase of the moon given its position and the suns position in the sky. Can someone...

What you mean to say is that I can cancel out the r's across the equation? m1r^2(a/r) = (m2(g-a))-m1a (r) So instead I would have: m1a = m2(g-a) - m1a

I'm beginning to think we are in the same school Torater.. (Macewan?) BTW, yeah I don't believe there would be an x direction because that would include friction which arises from a rolling sphere and in this case it's an unrolling condition. I'm happy we've made it through these problems...

I'll do mine now lol m1=6kg m2=3kg r=0.25m Alright so my three equations are: T=m2(a-g) F=T-m1a FR=m1r^2(a/r) Substituting: m1r^2(a/r) = (m2(g-a))-m1a (r) 6(0.25)^2(a/0.25) = (3(9.8-a))-6(a)(0.25) 1.5a = 29.4-3a-6a (0.25) 1.5a = -2.25a + 7.35 3.75a = 7.35 a= 1.96m/s^2...

Yeah yours looks correct, I'm just confused as to how you got this: 2/5m1r²(a/r) = (m2(g-a) - m1a) (.5) Can you explain how you got that please (negating the fact that the moment of inertia will be different)

Alright I isolated T and F: (1) T= -m2(a+g) (2) F=T-m1a and I have FR=mR^2 (a/r) leftover When I substitute either equation 1 or 2 into an equation I still have acceleration left over which is an unknown

From what you've said I got these equations: ∑Fy=m2a => T-m2g = -m2a (wouldn't tension for the hanging mass be positive, and the acceleration be negative?) ∑Fx=m1a => -F+T = m1a ∑t=Iα => FR=Iα => FR = mr2(a/R) From here I'm having trouble putting these equations together, where is a good...

Hmm I don't think the fR would be included with the rotating wheel because tension is pulling directly on the wheel whereas friction is not, so friction only comes into account on the x-axis.

The fR is the frictional force*radius (I don't know whether or not this should be included). I used static friction because the rolling sphere only makes contact at one point along the horizontal, but yeah it shouldn't make a difference if you're using kinetic friction. I'm having the same...

I am having trouble with almost the same problem (could you tell me the difference between these two?) A hoop shaped 6-kg wheel of radius 25cm is attached to a 3.00kg mass by a mass-less cord that passes over a frictionless, mass-less pulley. When the system is released the mass drops and the...