Recent content by hi im nimdA

So just to clarify, we are still using the equation that I solved for in part (a) and then since the force is attractive, we tack on a negative sign? If so, thank you for all of your help!

K_i+U_i=K_f+U_f Since it's initially at rest, -\frac{Gm_1m_2}{r_i} = \frac{1}{2}m_1v_f^2-\frac{Gm_1m_2}{r_f} -\frac{Gm_2}{r_i} +\frac{Gm_2}{r_f} = \frac{1}{2}v_f^2 v_f = \sqrt{2Gm_2(-\frac{1}{r_i}+\frac{1}{r_f})} v_f = \sqrt{2G(100)(-\frac{1}{1}+\frac{1}{0.5})} v_f=10\sqrt{2G} But when...

Ah. That's a good way to think about it. Thanks for correcting my misunderstanding.

Darn, I thought I could easily just apply it to its position instead of time. Mechanical energy conservation? Do you mean E_{mech, i} = E_{mech, f} K_i + U_i = K_f + U_f . We haven't covered energy for gravitation in lecture yet, but I found in the textbook that gravitational potential...

First, I started with F_a = m_aa_a=G \frac{m_am_b}{r^2} and F_b = m_ba_b=G \frac{m_am_b}{r^2} . Solving for their respective accelerations, I got a_a=G \frac{m_b}{r^2} = 100G and a_b=G \frac{m_a}{r^2} = 100G, meaning that the initial acceleration of the two point particles are each 100G ...

To start this problem, I used equation (1) K_i + U_i = K_f + U_f Then, using (2) and (3) and knowing that the initial velocity is 0, I have m_igy_i = \frac{1}{2} m_fv_f^2 + m_fgy_f The mass of the hanging part of the rope is ## \frac{y_0}{L} m ##. Additionally, I set the face of the table...

So for part (a), I used the fact that 1/N = the period = T. I solved for the velocity, where i got v=2πRN. I plugged that v into F=(m v^2)/R, which is the centripetal force, but also the force of friction. My answer to part (a) is 4(π^2)mR(N^2). I'm a bit confused on part (b). I know that...