Recent content by Nash77

So, we found a new way to solve for acceleration and it works, just not sure why. a = g(m1 - m2) --------------- (m1+m2+1/2MP) ^^ where MP = mass of the pulley. Why did we add in the 1/2 mass of the pulley to solve?

Please refer to problem VI here: http://www.physics.ohio-state.edu/~dongping/phys131_winter2011/phys131/final-exam-practice-2011.pdf Use 10 for gravity if needed. Homework Statement 1. When the system is released from rest, what is the magnitude of the blocks' acceleration? 2...

Thanks! Another equation that worked out was replacing h (0.05) for L(1-cos theta) from an energy equation PE= mgr(1-cos theta). Any idea where that equation comes from?

For a particular homework problem, we needed to find an angle theta. This equation was used: h = L(1 - cos theta) to solve for theta. I guess that came from the Kinetic Energy equation KE = mgR(1-cos theta), but where does that kinetic energy equation come from? Is it just an...

Homework Statement V. See the diagram attached --> problem V. A 1-kg mass rod has a length of 1 m and is attached to a vertical wall at point Q. The rod is free to rotate around Q. The other end of the rod (P) is tied with a string fixed to the wall. The system is at rest. Take g=10...

Thanks, I will do that right now. While on this one thought, I've been working on IV. (bullet and block) I found the momentum of the bullet, = mv = 0.01*100 = 1 N*s This momentum should be conserved when the bullet is lodged in the block, so the system has the same momentum, 1 N*s, so 1...

Attached is my practice physics final. I need help! Below are my attempts at solutions... probably way off but maybe I just need a head start. Thanks for looking! PS We are told to use 10 m/s^2 for gravity. --Nash. III. 1) at the top, all potential energy = mgh = 1*10*6 = 60...

Kinetic energy = 1/2*I*v^2, where v = angular speed. I = moment of inertia, and in this case I found the inertia of the hoop = m*R^2 + m*(3R)^2 = 10*m*R^2. The inertia of the rod = 1/3*m*L^2 = 1/3*m*(2R)^2. ^^ I understand the moment of inertia equations, I = mR^2, but why are we...

I understand the general equations in this problem, but why is the hoop being multiplied by (3R) [I = mR^2 + m(3R)^2] and the rod being multiplied by (1/3) [I = 1/3mL^2