Recent content by tseryan

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

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

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