# Acceleration of a solid cylinder on an incline

1. Sep 13, 2009

### chelseaalyssa

1. The problem statement, all variables and given/known data
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)

2. Relevant 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

3. 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:
Simplified:
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 ...

2. Sep 13, 2009

### JANm

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

3. Sep 13, 2009

### Astronuc

Staff Emeritus
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)