# Rotational Kinematics - Cylinder down a slope.

1. Nov 2, 2008

### CaptainSFS

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

A certain non-uniform but cylindrically symmetric cylinder has mass 9 kg, radius 1.2 m, and moment of inertia about the center of mass 7.6 kg m2. It rolls without slipping down a rough 20° incline.

What is the acceleration of the cylinder's center-of-mass?

2. Relevant equations

kinematics and the rotational equivalents

3. The attempt at a solution

The hints I was given was to find/use 3 equations. I believe the equations I found are wrong. I seem to have some difficulty with kinematics, and now especially this rotational kinematics.

I found:
m*g*sin(20) - F(friction) = m*a
T(orque) = I * alpha
a = alpha * r

Any help to whether these are the useful equations, and how I may go about using these? Thx. :)

2. Nov 2, 2008

### Staff: Mentor

What is the relationship between torque on the cylinder and the frictional force at the circumference?

3. Nov 2, 2008

### CaptainSFS

Are they the same?

b/c the frictional force is a force not coming out from the CM; therefore it helps it spin? I believe it is the only other force, so they must be the same?

Last edited: Nov 2, 2008
4. Nov 2, 2008

### CaptainSFS

I also tried using the gravitational force as the Torque.
I used m*g*sin(20)=I*alpha
m*g*sin(20)/I=alpha and alpha*R = a
so... m*g*sin(20)*R/I=a,
but this didn't work either.

-------------------
I just solved this problem.

Last edited: Nov 3, 2008
5. Nov 20, 2008

### llamaworm

Use these equations:
mgsin(20) - F(friction) = ma
I(alpha) = F(friction)*r
a=(alpha)*r

The only force that applies a torque is friction, because the normal force and gravity both act on the center of mass.