Recent content by calef

Exactly.

Remember, your ball is rolling without slipping. So you do have to include both the rotational and kinetic. The energy balance equation you wrote doesn't easily correspond to a physical situation, so imagine, instead, you wrote: mgh = (1/2)mv^2 This would correspond to a ball sliding down...

Almost right. What happens when you multiply an acceleration by a time? What are the units? (You have the correct numeric answer, your units are off)

1) Your formula for the inertia of a solid disk is incorrect. 2) What other energy is in your system that you have not yet accounted for?

And if you think about it, it makes sense that the Earth's translational kinetic energy is several orders of magnitude greater than its rotational. In fact, if you were only solving the net kinetic energy of the Earth to a couple decimal places, you could neglect the rotational kinetic energy...

I'm not exactly sure what they're asking here. For constant accelerations, you "relevant equation" is basically the answer, assuming you swap out the "x" for a "+" and take a square root. What's throwing me is the request for proof by chain rule.

Yeah, ignore that entirely. I misread what you wrote. You don't need to solve for the mass.

What's the mass of the hoop? (this is only relevant to check your work) (and with your work, I arrived at 5.21 seconds on the ramp)You can simplify this problem immensely for the second part. Just treat it like an object decelerating from 3.3 m/s to 0 m/s over 4.3 meters. Once you've solved...

I never thought about these problems in terms of tension--not at least until the pulley actually had friction acting on it. But yeah, T=0.

You get to find the right equation ^_^. The biggest hint I can give you is that F has the same acceleration as the force acting on m2.

That's another tricky part. There's no force "opposing" m2's motion. Imagine you're standing on an infinite plane of ice, and let's say your feet are frictionless against the surface. Suddenly you're being pulled along by a rope with magnitude, let's saaaay, m3g. So you're gliding along this...

Don't assume it's on ground level. The tricky thing about trajectory is that the angle which gives the farthest horizontal distance at any given velocity is below 45 degrees when you raise the launch platform. It's not much more difficult to solve as if it were on the ground, though. Just...

Think about this problem as if m3 wasn't there. Once F acts, m2 is, relative to the ground, stationary, and m1 will be accelerated to the right. Now, think about the problem as if m2 wasn't there. Once F acts, m1 and m3 are accelerated to the right, and m3 falls to the ground. Now, put both...

What direction do you think you need to take this problem?

I think (A) is correct, because Roger is the only external force in this problem. Without Roger, the diving board would not be deformed at all, and Roger's kinetic energy is only really changing because of what he did. The tricky wording here kind of gives it away: Ludmilla's claim doesn't...