I had a quick question on this one. To find point G i used (L/2 + L)/(m + 2m) = L/2. In words, the center of mass of a bar (L/2) plus the location of the particle (L) divided by the total mass. Is this the correct method to find G?
Subscribe to:
Post Comments (Atom)
5 comments:
There are a couple ways to work this problem.
One is to treat the rod and particle as a single body (as you are doing). Here you need to find the center of mass of the system. The location of the cm for the total system is (m*L/2 + 2m*L)/(m + 2m). With this approach, you calculate the PE based on the position of the cm for the total system. The mass moment of inertia about the fixed point O is IO = IO_bar + IO_particle = [(1/12)*m*L^2 + m*(L/4)^2] + [(2m)*(3L/4)^2]. Therefore, the KE is (1/2)*IO*omega^2.
The other approach is the treat the bar and particle individually, and add together in the work-energy equation. PE_bar is based on the cm of just the bar. PE_particle is based on the location of the particle. KE_bar = (1/2)*IO_bar*omega^2. KE_particle = (1/2)*(2m)*vP^2. You will use kinematics to say that vP = omega*(3L/4).
Thanks for the clarification
I may be wrong, but when I used KE-bar = (1/2)*IO_bar*omega^2, I got the wrong answer. When I used KE_bar = (1/2)*IO_bar*omega^2 + (1/2)*(m)*vB^2, I got the right answer. It may have just been a miscalculation, but this actually makes more sense to me...
Using this secound approach, would the equation then be:
V_1_rod + V_1_ball + T_1_rod + T_1_ball...etc
--to Jeff--
I was not clear in my second approach above. The TOTAL kinetic energy for the system is:
KE_total = KE_bar + KE_particle = (1/2)*IO_bar*omega^2 + (1/2)*(2m)*vB^2.
This is what you have used (except you have "m" rather than "2m" for the particle).
Are we saying the same thing??
Post a Comment