Conservation of Energy and the Race between Different Shapes on an Inclined Ramp

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SUMMARY

The discussion focuses on calculating the linear speeds of a ring, disk, and sphere released from a 3.80-meter inclined ramp at an angle of 17.0° using the principle of conservation of energy. The relevant equations for the moment of inertia are provided: Iring = mr², Idisk = (mr²)/2, and Isphere = (2mr²)/5. The conservation of energy equation is applied, leading to the relationship mgh = (1/2)Iω² + (1/2)mv². The user successfully simplifies the equation by substituting the moment of inertia for the ring and recognizes the importance of correctly applying the angular velocity in the calculations.

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Homework Statement


Three uniform objects: a ring, a disk, and a sphere all have identical masses and radii. They are released simultaneously from the top of a ramp 3.80 meters in LENGTH. If the ramp is inclined at θ= 17.0°, use conservation of energy to calculate the linear speed of the ring, disk and sphere.


Homework Equations


Ei=Ef
Iring = mr2
Idisk = (mr2)/2
Isphere = (2mr2)/5


The Attempt at a Solution


Ui + Ki = Uf + Kf
mgh = (1/2)Iω2f + (1/)mv2f <--initial K and final U are zero

then substituting I for Iring
mgh = (1/2)((mr2)(vf)/(r)) + (mv2f)
masses cancel, pulled out 1/2
gh = (1/2)(rvf + v2f)

But that's as far as I can get before I can't understand it.
 
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robertmatthew said:

The Attempt at a Solution


Ui + Ki = Uf + Kf
mgh = (1/2)Iω2f + (1/)mv2f <--initial K and final U are zero

then substituting I for Iring
mgh = (1/2)((mr2)(vf)/(r)) + (mv2f)

Note that ωf is squared.
 
I always end up asking questions on here because of stupid mistakes like that, haha. That made much more sense, with the radii canceling out. Thanks so much.
 

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