your angular acceleration looks like you took the second angular velocity and divided it by the entire time of motion.
Realize that angular acceleration can only occur if there is torque about that axis. The problem here is that you have two sources of torque: 1. applied force 2. friction...
good thinking, but I think you may be using the wrong approach. Think about what the use of the word "equilibrium" implies about the sum of the FORCES. The way I read that question is it is asking you the angle the springs make once the entire system has come to rest (i.e. no longer bouncing...
for anyone who looked at this and is curious, here is an update.
I thought this would be a straightforward dynamics problem given all the assumptions; however, it turns out to be something beyond the scope of what is solvable in a beginning dynamics course. The reason is the variable moment...
I think you have to treat each body separately. you have a net loss of energy since your coefficient of restitution is less than 1. You should break up the collision into 3 parts. Find the energy before, the change during, and the energy after the collision.
Check this out, it should be of...
More generally, the integral of the sum of all the forces acting on the body over time is equal to the net change in momentum. This can be broken down into component vectors.
integral of sigma Fdt = delta G
Likewise, the integral of all the moments applied over time is equal to the net...
mv + integral of F dt from t=0 to t=t is equal to mv final. Since V(bp1y) is zero and V(bp2y) is known, V(bp2y) must equal the vertical component of impulse (integral of Fdt).
Then use H(A1) + integral of Mdt from t=0 to t equals H(A2). H(A1) = 0 since it isn't rotating, so integral Mdt =...
If you have V(bpy), that should be equal to and opposite the impulse applied to point p on A. Use the principle of impulse and momentum to find omega(A)
OK, I'm thinking that the I.C.R. must be perpindicular to both velocity vectors for the body correct? If you look at the direction of the velocity of P on A just after the collision, there is only one way to draw a perpindicular line to that vector so that you could draw another perpindicular...
right. Have you tried using the relative velocity equations?
V(A2) = V(P2) + V(P/A2)
V(P/A2) = r (P/A) X omega (P/A)
since they are rigid bodies, then omega (P/A) should be the same omega for the entire body...
If you have rotation, you always have a normal component of acceleration (centripital acceleration). If there is no moment or other external forces applied to the system, then the net acceleration of any point in the system other than the centroid is just the cross product of omega and the...
One approach you may consider:
omega(A) is related to Vay through the acceleration vector.
acceleration = a(tangential) + a(normal)
a(tangential) = r X alpha
a(normal) = omega X (omega X r) = omega X V(tangential)
Make sure you take all the rotations about the center of mass.
Find...
I went back through the problem, and found a few careless mistakes. I now know which way to go, but I cant figure out how to get around this differential equation:
(alpha) = (omega)' = (theta)'' = 5.43sin(theta) - 43.19cos(theta)
If someone could help me solve that for theta, then I think...
So plugging everything back into my energy equation, I get the following:
maG(x)*x + 181.3 = maG*x + maG*x + 1.22mg
Next, I need to find the acceleration components of G. Using aG = ao + aG/O, I can easily see that ao is just the friction force divided by the mass. This yields 4.05 m/s2...
Homework Statement
Background:
This is a problem I made up as part of an extra credit assignment for my first dynamics (engineering) class. I know the system will of course roll over, and he will end up looking like a fool, but that is besides the point. Keep in mind also that I can modify...