Recent content by kelslee28

It's right.

So this is what I did. m1 = mass of ball1 and ball2 m2 = mass of putty 1/2Lm2V = Iw (of the ball on the left) + Iw (the ball on the right plus the mass of the putty) 1/2Lm2V = [m1((1/2)L)2 + (m1+m2)((1/2)L)2]w solve for w [1/2Lm2V]/[m1((1/2)L)2 + m1+m2((1/2)L)2] I got 0.233 rad/s Edit Is...

I can't use the constant acceleration equation because it's not in freefall. What I needed to do, which now I have figured out, is solve for v using the fact that mg(R-r) = 1/2mv^2 + 1/2Iw^2 where w = v/r Then use that V to solve for the rotational kinetic energy. Divide this by the total...

Homework Statement Two balls of mass 2.26 kg are attached to the ends of a thin rod of negligible mass and length 72 cm. The rod is free to rotate without friction about a horizontal axis through its center. A putty wad of mass 145 g drops onto one of the balls, with a speed 2.7 m/s, and sticks...

Homework Statement A small sphere, with radius 1.6 cm and mass 5.2 kg, rolls without slipping on the inside of a large fixed hemispherical bowl with radius 0.82 m and a vertical axis of symmetry. It starts at the top from rest. What is the kinetic energy of the sphere at the bottom? What...

Yes, I got it! Thanks a lot. I'm home sick with the flu so this means a lot that someone would help me.

So the top and bottom would be I= 1/3 mL2 but it's only a portion of the mass, right? The right side would be I = mh2 and the h would be the length b. The only thing I can think of for the left side is the formula for a solid cylinder, but for that you need a radius.

Homework Statement A thin uniform wire is bent into a rectangle. The short, vertical sides are of length a, and the long, horizontal sides are of length b. If the total mass is 31.00 grams, a = 25.00 cm and b = 44.30 cm, what is the moment of rotational inertia about an axis through one of the...