Rolling Cylinder Forces

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Homework Statement



A cylinder with moment of inertia I, mass M, and radius r is being pulled by a string with a tension T.

What is the acceleration of the cylinder? What is the force of friction?


Homework Equations



Fnet = M*a = T - f
Torque = I*(a/r) = f*r

The Attempt at a Solution



So a = f*r^2/I
And M*(f*r^2/I) = T- f.

What I don't understand is how "f" (friction) slows the cylinder. It is just causing rotation, not acting on the center of mass. I thought my professor said that rolling friction doesn't slow things down...

Thanks.
 

Answers and Replies

  • #2
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135

Homework Statement



A cylinder with moment of inertia I, mass M, and radius r is being pulled by a string with a tension T.

What is the acceleration of the cylinder? What is the force of friction?


Homework Equations



Fnet = M*a = T - f
Torque = I*(a/r) = f*r

The Attempt at a Solution



So a = f*r^2/I
And M*(f*r^2/I) = T- f.

What I don't understand is how "f" (friction) slows the cylinder. It is just causing rotation, not acting on the center of mass. I thought my professor said that rolling friction doesn't slow things down...

Thanks.
The acceleration 'a' in ƩF = Ma is the acceleration of the COM(Center of Mass) .ƩF is the net force that acts on the body .The COM is a special point which moves as if all the forces acting on the body were acting on it,irrespective of the point of application of the force .

Friction causes a torque about the COM ,but at the same time produces acceleration of the COM.

It is not clear from the OP where is the point of application of Tension ?
 
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  • #3
haruspex
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What I don't understand is how "f" (friction) slows the cylinder. It is just causing rotation, not acting on the center of mass. I thought my professor said that rolling friction doesn't slow things down...
I expect your Prof said (or should have said) that rolling friction does not reduce the mechanical energy. What it does do is arrange that a portion of the KE is in rotational form. That therefore reduces the KE in linear form, i.e. slows it down.
If the centre of the cylinder is being pulled horizontally with force T, the net horizontal force is T-F, the torque about the centre is Fr:
ma = T-F
Iα = Fr
a = rα
Ia = Fr2 = (T-ma)r2
a = Tr2/(I+mr2)
If I = mr2/2:
a = 2T/3m
An alternative approach is to take moments about the point of contact with the ground:
(I+mr2)α = Tr
That gets to the answer a little faster.
 

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