Recent content by volleygirl292

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    Using Lagragian equations to find accelerations

    Right so I defined the v in 1/4 to be x2' since it is moving with the block and then then v in 1/2(m3)x3'^2 so since x2' and x1' are equivalent I replaced them in the L to be x' and then left x3' so my L=(1/2)(x')^2 (m1+m2+(1/4)m3)+(1/2)m3(x3')+m2g(l-∏a-x2)
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    Using Lagragian equations to find accelerations

    So for the added KE of cylinder due to the ground would be (1/2)mv^2? Also, I was wondering if I need to deal with the fact that the cylinder is on m1 so it is a distance above the table which is equivalent to the height of that m1.
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    Using Lagragian equations to find accelerations

    Also, I do not need to add anything to the potential energy right?
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    Using Lagragian equations to find accelerations

    Yes, I did mean T not I. I guess I am just confused about what all I have to add to the KE. I thought that the ((1/4)mv^2) was the term due to the KE associated with the movement of the centre of mass.
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    A thin vertical rod sliding until it hits a lump of putty and rotates.

    So ∫ (r x mv dr) where the integral is from 0 to d? I am not sure how to integrate that then. I am just very confused.
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    A thin vertical rod sliding until it hits a lump of putty and rotates.

    Can I just use the center of mass so I do not have to do all those integrals? Our professor said the math wouldn't be very long on this problem.
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    A thin vertical rod sliding until it hits a lump of putty and rotates.

    Also, I do not understand conceptually how the rod can have angular momentum before it rotates. I thought to calculate angular momentum the object must be rotating.
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    A thin vertical rod sliding until it hits a lump of putty and rotates.

    Right, so L=Ʃri x (mi)(vi) and Ʃmi=m and v is the same for all the elements and Ʃr=d (length of the stick so why is L=mvd incorrect for before the rod hits the putty?
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    A thin vertical rod sliding until it hits a lump of putty and rotates.

    so L=mvd before and mvdsinθ after?
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    A thin vertical rod sliding until it hits a lump of putty and rotates.

    oh right thanks!
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    A thin vertical rod sliding until it hits a lump of putty and rotates.

    The definition for a particle is L=mvrsinθ and θ=0 if it is moving in the horizontal direction and not rotating yet.
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    A thin vertical rod sliding until it hits a lump of putty and rotates.

    Homework Statement A thin vertical rod of mass m and length d slides on a frictionless horizontal surface with velocity Vo to the right. It his and sticks to a small lump of putty and rotates about the putty at point O. (a) Find the angular momentum of the rod about O just before and just...
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    Using Lagragian equations to find accelerations

    Homework Statement A mass m1 slides on a frictionless horizontal table. it is attached by a massless cord passing over a massless pulley to a mass, m2. A cylinder of mass m3, radius r, and moment of inertia 1/2(m3*r^2) rests on m1. (a) Choose and specify generalized coordinates (two are...
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    Exact length of the curve analytically

    I got it down to the integral of sqrt of 1+ (27-x)/x but i still have an x in the denominator
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    Exact length of the curve analytically

    ok so now i have it down to integral of sqrt of 1-(9-x^(2/3)/(x^(2/3)) but i still have a exponent in the denominator