Rolling without slipping and a string

[trulyfalse]

Hello PF!

Homework Statement

A string is wrapped around the small cylinder as shown. You pull with a force F (and the cylinder does not slide). Calculate the acceleration of the cylinder CM (including direction). Here r = 3 cm,
R = 5 cm, F = 0.1 N, and m = 1kg. [Make sure to define one direction of rotation (CW or CCW) as positive, just like you define one direction of X as positive. If you use a relationship like v=R
or a=R you need to make sure that your definitions are consistent]

Correct solution: 0.0267 m/s^2

Homework Equations

Sum of the torques = Iα
Moment of inertia of a solid cylinder = 1/2MR^2

The Attempt at a Solution

Let counter clockwise torques be positive for the purposes of this solution.

I started by calculating the torques about point C, where C is the center of mass of the cylinder. Since the cylinder is rolling without slipping, the force due to static friction on the cylinder must be equal and opposite in magnitude to the applied force, F.

Ʃt = R*Fs - r*F
ƩF = Fs - F = 0
Iα = R*F - r*F
ac/R = (R*F - r*F)/I
ac = 2(R*F-r*F)/(MR)
ac = 0.08 m/s^2

Is it wrong to equate Fs to F in this case? Or is there another factor that I'm not considering?

 

[ehild]

Hello PF!

Homework Statement

A string is wrapped around the small cylinder as shown. You pull with a force F (and the cylinder does not slide). Calculate the acceleration of the cylinder CM (including direction). Here r = 3 cm,
R = 5 cm, F = 0.1 N, and m = 1kg. [Make sure to define one direction of rotation (CW or CCW) as positive, just like you define one direction of X as positive. If you use a relationship like v=R
or a=R you need to make sure that your definitions are consistent]

Correct solution: 0.0267 m/s^2

Homework Equations

Sum of the torques = Iα
Moment of inertia of a solid cylinder = 1/2MR^2

The Attempt at a Solution

Let counter clockwise torques be positive for the purposes of this solution.

I started by calculating the torques about point C, where C is the center of mass of the cylinder. Since the cylinder is rolling without slipping, the force due to static friction on the cylinder must be equal and opposite in magnitude to the applied force, F.

Ʃt = R*Fs - r*F
ƩF = Fs - F = 0
Iα = R*F - r*F
ac/R = (R*F - r*F)/I
ac = 2(R*F-r*F)/(MR)
ac = 0.08 m/s^2

Is it wrong to equate Fs to F in this case? Or is there another factor that I'm not considering?

The cylinder is accelerating and you need to find the acceleration of its CM. The acceleration of the CM is equl to the resultant force over the mass. The resultant force is F-Fs, why should it be zero?

ehild

 

[trulyfalse]

Oh jeez, it seems I made quite an egregious error. I initially assumed that the sum of the forces would be equal to zero on the object since static friction is involved, but after reading your comment and thinking about it further I realized that's only because the velocity of the instantaneous axis is equal to 0. Thanks ehild. I'll calculate the torques about that axis instead and see if I get the correct answer.