Recent content by kingsmaug

I don't believe I'm using the wrong ones. So working it over using the proper masses: Homework Equations KE = 1/2mv^2 KE = 1/2Iw^2 L=mvr L=Iw I=ML^2/3 I=mr^2 r=L (I'm using the pivot as the point of origin) mvL=(ML^2/3+mL^2)w w=mvL/(ML^2/3+mL^2) = 1.47059 rad/sec (I don't understand why it...

Homework Statement A piece of putty of mass m = 0.75 kg and velocity v = 2.5 m/s moves on a horizontal frictionless surface. It collides with and sticks to a rod of mass M = 2 kg and length L = 0.9 m which pivots about a fixed vertical axis at the opposite end of the rod as shown. What fraction...

Units, yes. But I just realized that I left out the r^2 term from I. So the rotational energy is 1/2*I*w^2, and w=v/r That turns to 1/2*I*(v/r)^2 But I suppose it doesn't entirely matter anymore. I've run out of attempts on the problem.

*facepalm* Energy loss due to friction. So, KEi = KEf + Ertof - Efric 1/2mvi^2=1/2mvf^2+1/2(2/5m)(vf^2/r^2) - uk(N)(d) Which when solved gives me vf=1.59 m/s, which is also wrong.

Homework Statement A spherical bowling ball with mass m = 4.3 kg and radius R = 0.102 m is thrown down the lane with an initial speed of v = 8.2 m/s. The coefficient of kinetic friction between the sliding ball and the ground is μ = 0.29. Once the ball begins to roll without slipping it moves...

My guess would be to include electrostatic potential energy. So, if Q=q, then U=kQq/r becomes U=k/r, and if r is very small, and some arbitrary value 1, then U=1. So, very close the total energy would be KE+U, but then the U would just cancel out. Well, if you look at it as a regular system...

Homework Statement A proton of mass m is moving with initial speed v0 directly toward the center of a nucleus of mass 31m, which is initially at rest. Because both carry positive electrical charge, they repel each other. Find the speed v' of the nucleus for the following conditions: a) the...