Recent content by tseryan

1. Rotational Inertia (Moment of Inertia) of a Rod

Got it! Thanks Doc Al! Now I don't understand why there is a Parallel Axis Theorem for a rod with the equation I=(1/3)M(L^2). What would that be used for?
2. Rotational Inertia (Moment of Inertia) of a Rod

It is I=MR^2. If I find the sum of the point masses of inertia I get the moment of inertia? Does it matter that it's on a rod?
3. Rotational Inertia (Moment of Inertia) of a Rod

Homework Statement 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. Homework Equations...
4. Moment of inertia of a thin spherical shell

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: Icm=(2/3)M(R^2) The "cm" is in subscript by the way. Icm is...
5. Conservation of Momentum in an Inelastic Collision

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.
6. Conservation of Momentum in an Inelastic Collision

Ahhh! I should've caught that mistake. I now have Va=sqrt(4gL). I understand how to get v, but I am still unclear why mv/2M and sqrt(4gL) are equal? Are they just two different ways to write Va?
7. Conservation of Momentum in an Inelastic Collision

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?
8. Conservation of Momentum in an Inelastic Collision

I don't quite understand. Is that the 1/2mv^2 equation?
9. Conservation of Momentum in an Inelastic Collision

Correct, that is what I have. I solved for Va and got Va=mv/2M. I'm not sure what to do after this. How would I find Vb and v from this?
10. Conservation of Momentum in an Inelastic Collision

Ahhhh, I simply solve for the variable Va. Let me try that! :) I really appreciate your help Rainbow Child.
11. Conservation of Momentum in an Inelastic Collision

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.
12. Conservation of Momentum in an Inelastic Collision

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.
13. Conservation of Momentum in an Inelastic Collision

Are you sure? I understand a=0 at that point, but the pendulum is moving the entire rotation.
14. Conservation of Momentum in an Inelastic Collision

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...
15. Conservation of Momentum in an Inelastic Collision

It means that the pendulum swings around in an perfect circle (it stops exactly where it starts).