Determine the translational speed of the cylinder

AI Thread Summary
The discussion focuses on determining the translational speed of a solid cylinder rolling down an inclined plane. The approach involves equating gravitational potential energy (mgh) to the sum of translational and rotational kinetic energies. The derived formula for the translational speed at the bottom of the incline is v = √[(4/3)gh]. Participants confirm that this method is correct and suggest an alternative approach involving effective mass and gravity, although it may not be favored by examiners. The solution emphasizes the relationship between potential energy and kinetic energy in rolling motion.
k-rod AP 2010
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Homework Statement


A solid cylinder w/ mass m, radius r, and rotational inertia 1/25mr2 rolls without slipping down an inclined plane. THe cylinder starts form rest at height h. The inclined plane makes an angle θ w/ the horizontal.Determine the translational speed of the cylinder when it reaches the bottom of the inclined plane.


Homework Equations


v=velocity
ω=angular speed
I=moment of inertia
U=gravitational potential energy
K=Kinetic Energy

The Attempt at a Solution


U=Ktranslational + Krotational

mgh=(1/2mv2) + (1/2Iω2)

mgh=(1/2mv2) + (1/2 (1/2mr2)(v2/r2))

gh=(v2/2) + (v2/4)

gh=(2v2/4) + (v2/4)

gh= (4/3)v2

v2=(4/3)gh

v= √[(4/3)gh] would this be the correct procedure to solve this?
 
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hi k-rod AP 2010! :smile:
k-rod AP 2010 said:
A solid cylinder w/ mass m, radius r, and rotational inertia 1/25mr2 rolls without slipping down an inclined plane. THe cylinder starts form rest at height h. The inclined plane makes an angle θ w/ the horizontal.Determine the translational speed of the cylinder when it reaches the bottom of the inclined plane.

v2=(4/3)gh

v= √[(4/3)gh] would this be the correct procedure to solve this?

(i assume you meant rotational inertia 1/2 mr2 ? :wink:)

yes, that's the right method and result! :smile:

(btw, another way is to say the "rolling mass" is I/r2, = m/2, so the total "effective mass" is 3m/2, so the "effective gravity" is 2g/3, so 1/2 mv2 = 2mgh/3 … but i don't think the examiners would like that! :redface:)
 
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