Inclined plane translational velocity

AI Thread Summary
A hollow cylinder with a mass of 1.2 kg and a diameter of 0.25 m rolls down a 35-degree inclined plane, traveling 3 meters. The height calculated from the incline is approximately 1.72 m, leading to a translational velocity of about 4.1 m/s using the equation v = √(gh). The discussion emphasizes the conservation of energy principle, equating potential energy at the top with kinetic energy at the bottom. The relationship between translational and angular velocity is also highlighted, with the final equation confirming that the calculated velocity is correct. The key takeaway is the effective application of energy conservation to determine the translational velocity of the cylinder.
davidelete
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Homework Statement


A hollow cylinder with a mass of 1.2 kilograms and a diameter of .25 m rolls down an inclined plane angled at 35 degrees. What is its translational velocity at the bottom of the incline plane if the distance traveled is 3 meters?

Homework Equations


v=\sqrt{gh}
?

The Attempt at a Solution


sin35=h/3m
h=1.72m
v=\sqrt{9.8*1.72}
4.1 m/s
 
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According to the conservation of energy,
mgh = 1/2*Iω^2 + 1/2*m*v^2.
here h = d*sinθ. The angular velocity v =ω*R, where R is the radius of the cylinder.
 
Last edited:
rl.bhat said:
According to the conservation of energy,
mgh = 1/2*Iω^2 + 1/2*m*v^2.
here h = d*sinθ. The angular velocity ω = v*R, where R is the radius of the cylinder.

I need translational velocity.
 
davidelete said:
I need translational velocity.
v is the translational velocity of the center of mass.
 
rl.bhat said:
v is the translational velocity of the center of mass.

Please explain more.
 
According to the conservation of energy
K1 + U1 = K2 + U2
0 + mgh = 1/2*m*v^2 + 1/2I*ω^2.
mgdsinθ = 1/2*m*v^2 + 1/2MR^2(v/R)^2 + 0
So mgdsinθ = mv^2.
So what ever you have calculated is the required result.
 
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