Rotational Kinematics - Cylinder down a slope.

AI Thread Summary
A non-uniform cylinder with a mass of 9 kg and radius of 1.2 m rolls down a 20° incline without slipping, prompting a discussion on calculating its center-of-mass acceleration. Key equations identified include m*g*sin(20) - F(friction) = m*a, I*alpha = F(friction)*r, and a = alpha*r. The frictional force is the sole contributor to torque, as both the normal force and gravitational force act at the center of mass. The initial attempts to solve the problem involved confusion over the equations, but the correct relationships were ultimately clarified. The discussion highlights the importance of understanding the role of friction in rotational motion.
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Homework Statement



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?


Homework Equations



kinematics and the rotational equivalents

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