Acceleration of a solid cylinder on an incline

[chelseaalyssa]

Homework Statement

Show that the acceleration of a solid cylinder rolling down an incline that makes an angle (theta) with the horizontal is given by the equation:
a=2/3 g sin(theta)

Homework Equations

Using the conservation of mechanical energy:
1/2 Mv² + 1/2 I w² + Mgh = 1/2Mv² + 1/2 Iw² + Mgh
***Note: w = angular velocity, I'm just not sure how to insert the symbol
v² = v²₀ + 2a(x-x₀)
I=1/2MR²

The Attempt at a Solution

After simplifying the conservation of mechanical energy equation and substituting in the equation for moment of inertia, I got:
1/4v² + gh = 1/2v² + 1/4v²

In order to solve for acceleration, I substituted in the equation v²=2a(d) ... where d is equal to distance down the incline:
1/4(2ad) + gh = 1/2(2ad) + 1/4(2ad)
Simplified:
gh = ad
Therefore, a = gh/d
And height/distance is equal to sin(theta):
a=gsin(theta)

The problem is that I don't know where the 2/3 comes in ...

Thankyou in advance for your help :)

 

[JANm]

v² = v²₀ + 2a(x-x₀)

Hello Chelseaalyssa
Nice problem you have there. I see you found that if you flip the cilinder to the side the acceleration would fully go to sliding. The 2/3 factor is one to prove. Part of the potential energy comes into rolling the cilinder. I have tried to find if the moment of inertia is indeed
I=MR^2/2. Have not found the page(s) in my mechanic book where this very nice way of using integrals is directed to the most common forms... Suppose that this is correct:
The most important thing is to connect the rolling to the moving; so the omega w and the velocity v.
Succes and greetings from Janm

 

[Astronuc]

In this equation, 1/4v² + gh = 1/2v² + 1/4v², the v on the left side is not the same as the v on the right side.

If v on the left is initially zero, i.e. the cylinder is at rest, then the equation becomes

gh = 1/2v² + 1/4v²

and then using v²=2a(d)

gh = 1/2(2ad) + 1/4(2ad).

 

[chelseaalyssa]

Oh ok - I see where I went wrong...
Thanks for the help!