Recent content by pixelized

ok it's just an energy problem. The difference in potential energy between the surface and the point it asks for is the kinetic energy which is 1/2 mv^2 Now you know that F = GMm/r^2 and Mass = p * 4/3 * pi * r^3 the mass of the object cancels so you don't need it and potential energy...

I tried I = (MR^2)/2 + MR^2 = (3/2)MR^2 T = 2pi * Sqrt(3/2 MR^2/MgR) = 2pi * Sqrt((3/2)r/g)) This didn't work is there something I missed?

yes that's what I used for part two. I tried 0 but it was wrong. So maybe they can rotate counter clockwise. How would that work?

I solved for 2 correctly but I can't figure out what to do for #3 Any help would be much appreciated. I think it might involve L = r x p , but I'm not entirely sure. 2. [1pt] Two lightweight rods L = 20.5 cm in length are mounted perpendicular to a vertical axle and at 180° to each other...

On a frictionless table, a glob of clay of mass 0.740 kg strikes a bar of mass 1.740 kg perpendicularly at a point 0.140 m from the center of the bar and sticks to it. If the bar is 0.660 m long and the clay is moving at 9.600 m/s before the impact, what is the final speed of the center of mass...

Nevermind I got it. Seems I needed to set one of the velocities as negative since it was an inelastic collision.

could someone help me with this problem? A 139 kg tackler moving at 2.53 m/s meets head-on (and tackles) a 87.4 kg halfback moving at 5.14 m/s. What will be their mutual speed immediately after the collision? Oh right. Here's what I've tried doing but no luck. m1 = 139kg V1 = 2.53 m/s...