A very light (meaning don't consider mass of the rod) rod is placed along the x axis. It has a mass m1=2.0kg at x=0, a mass m2=1.50kg at x=50cm, and a mass m3=3.0kg at x=100cm.
Find the moment of inertia of the system about a pivot point at x=0.
The parallel axis theorem helps in calculating moments of inertia of homogeneous rigid bodies with strange geometries. The general equation is:
I (moment of inertia) = Icm + MD^2
In the case of a thin spherical shell it is:
The "cm" is in subscript by the way. Icm is...
Ah, everything finally makes sense now. I re-read everything you've said and it's all very clear now. :)
Thanks again Rainbow Child! You're a physics genius! I extremely appreciate you sticking with me the 3 or so hours it took for me to complete this fairly simple problem.
I just solved for Va and got sqrt(2gL) (the mass canceled out). This equation gives me the speed of if the pendulum before the collision. I also have the speed of the pendulum after collision, which I found earlier from the conservation of momentum equation. Is this correct?
Doesn't Va of M is equal to zero since the pendulum is initially not moving? I don't quite understand what to do when the initial and final velocities of an object is zero, but it moves in between that period.
Ohhh, I'm sorry if I incorrectly explained the problem, but the bullet is traveling at v and it asks for the minimum speed of the bullet as it passes through the pendulum in order for the pendulum to go around one rotation.
I don't think it uses the equation I listed above with a set velocity. They give me L (the radius of the spin), which leads me to believe that I need to calculate the work done by the pendulum in the spin (W=FD and d is 2Lpi (circumference), but I'm not sure about force) and correlate the work...