Recent content by jcruise322

The puck is lying along the y-axis with its COM at the origin. The puck travels a distance above or below the y-axis in a straight horizontal line towards the rod

Then that would mean angular momentum would not be conserved. I realize that the stick's angular momentum is zero and the puck has angular momentum RELATIVE to the stick. The only angular momentum is relative angular momentum of the system before the collision which can be quantified as rxmv for...

?? Angular momentum around the impact point is nonzero. The system rotates around its center of mass. Both of them individually are zero in regard to angular momentum before the collision; they are non spinning.

Homework Statement Is it possible for the respective angular momenta of each individual particle in a system to be zero, but the system's collective angular momentum be nonzero? For example, a puck on a frictionless air table moves (without spinning) toward a point on a rod that is not the...

Aha, that is what I suspected. By the way Haruspex, you have commented on all of the posts that I have made on physics forums so far. Appreciate the help!

Homework Statement A sphere of radius .06 m and mass .5 kg rolls down a ramp that is angled 30 degrees down the incline. It starts rolling from a height of 7 feet and does not slip What is its final linear velocity? Now, I used mgh=translational +rotational KE and found that the final velocity...

Oh, I see now...there are three vertical forces acting on the object. I forgot about the vertical y component of the pull. Thanks, SteamKing. And, it should be uk(mg-T*sin(theta)). not T*cos(theta). Appreciate it.

Homework Statement A block of mass "m" is pulled across a rough table by a string at an angle "θ" above the horizontal at a constant velocity. What is an equation for the magnitude of the kinetic friction force, fk? The answer that I was given was μk(mg-T*cos(θ)) Homework Equations Fk=Fn*uk...

Using mgh=.5*m*v^2 vf=1400 m/s. Pretty close to 1390.1 :)

Sorry, my final equation is: -(GM1)/r1=.5*v^2-G(M1)/r2

Yo, I think I got this figured out. I would use energy conservation. Ei=Efinal. The satellite only has potential energy when it is stopped given by-(G*m1*m2)/(r1). r1 equals radius of Earth + altitude. If we ignore drag, part of this PE is converted into KE, .5*m*v^2, and the other goes into...

Exactly! Did you read all of what I said? I said that it didn't. Anyway, if you think that I am wrong, please give me your opinion. I have done all that I can do for now.

*Steamking...xD

100,000 meters NOT feet! Sorry, I made a transcription error! Yup, this baby is in space, Streamking!

So...I will try to think about this in the most logical way, I guess. So, Earth and the satellite are part of the system. I am not sure if we can treat gravity's acceleration as equaling 9.81 m/s/s here...can we just use 9.8 as an approximation? Anyway, the satellite has PE at 100,000 meters...