How Does Friction Affect the Acceleration of a Rolling Cylinder?

[John O' Meara]

A rope is wound around a cylinder of mass 4kg and with moment of inertia I=0.020 kg.m^2 about an axis along the cylinder axis, see attachment. If the cylinder rolls without slipping, (a)what is the linear acceleration of its mass center? (b)Repeat for the case where no friction force exists between the table and the cylinder.
So we have a force of 20N applied in the direction shown in the attachment, and the radius of the cylinder is r=10cm.
My attempt: (a)torque T=I*(alpha) where (alpha) is angular acceleration, therefore (alpha)=T/I = 20*.1/.02 = 100rad/s^2
a_tangential=r*(alpha) = 10m/s^2.
a_center=?
(b) F=m*a therefore a_center=20/4=5m/s^2.
Any help would be welcome in redard to what a_center is in part (a). Thanks.

 

[John O' Meara]

The attachment didn't work. The force of 20N is applied at the top of the rolling cylinder and parallel to the table surface.

 

[OlderDan]

You have not considered all of the forces acting in part a)

 

[John O' Meara]

Ok I forgot the frictional force f.
20-f=m*a
T-Tf=I*(alpha), where T= torque, Tf =friction torque, alpha= angular acceleration. Now
T-Tf=I*a/r; r*F - r*f = I*a/r therefore
F - f=I*a/r^2 ...(i)
-20 + f=-m*a ...(ii) add i & ii therefore
F-20=I*a/r^2-m*a; => a=20/6=3.333 m/s/s...as F=m*a
a_tangential=3.333m/s^2
a_center=. Can anyone tell me where I might be gone wrong and where?Thanks.

 

[OlderDan]

Can anyone tell me where I might be gone wrong and where?

There is something wrong with your torque. Look at your free body diagram.

 

[John O' Meara]

Are you saying : T-Tf=I*(alpha) is wrong? which I think you are. Please specify.Thanks. I have got: 20 - f=m*a; T=I*(alpha) and f=(mu)*n ...n=m*g. I have got two unknowns a and f (and really only one independent equation). I must be able to eliminate the frictional force f.

 

[OlderDan]

Are you saying : T-Tf=I*(alpha) is wrong? which I think you are. Please specify.Thanks. I have got: 20 - f=m*a; T=I*(alpha) and f=(mu)*n ...n=m*g. I have got two unknowns a and f (and really only one independent equation). I must be able to eliminate the frictional force f.

In your equation for ma you have tension and friction acting in opposite directions. Look at where these forces act on the cylinder and the directions. What does that tell you about the torques they produce?

 

[John O' Meara]

I think I have got what you are saying:
20 - f = m*a;
T + Tf =I*(alpha); therefore r*F + r*f = I*a/r and F+f=I*a/r/r; which gives a new equation 20 + F = I*a/r^2 + m*a, now what is F equal to? Is it equal to m*a if it is then a_tangential = 10m/s^2

 

[OlderDan]

I think I have got what you are saying:
20 - f = m*a;
T + Tf =I*(alpha); therefore r*F + r*f = I*a/r and F+f=I*a/r/r; which gives a new equation 20 + f = I*a/r^2 + m*a, now what is f equal to? Is it equal to m*a if it is then a_tangential = 10m/s^2

Those should be small f in your quote, right? I changed them. This gives you two equations for two unknowns, f and a. f is whatever it has to be to satisfy both equations, as is a.

 

[John O' Meara]

20 + f=I*a/r^2 ...(i); 20 - f = m*a ...(ii); a=40/(I/r^2+m) therefore a_tangential = 6.667m/s/s.The question is what is the linear acceleration of the center of the cylinder? The question then is what is the relationship of the linear acceleration to the tangential acceleration. a_tangential = r*(alpha) where (alpha) is the angular acceleration.

 

[OlderDan]

20 + f=I*a/r^2 ...(i); 20 - f = m*a ...(ii); a=40/(I/r^2+m) therefore a_tangential = 6.667m/s/s.The question is what is the linear acceleration of the center of the cylinder? The question then is what is the relationship of the linear acceleration to the tangential acceleration. a_tangential = r*(alpha) where (alpha) is the angular acceleration.

The (a) you defined earlier is the acceleration of the center of the cylinder. You then wrote your angular acceleration in terms of that (a) as alpha = a/r. The tangential acceleration relative to the center of the cylinder is (alpha)*r = = (a/r)*r = a, the same as the acceleration of the center.