F = MA Exam 2011 # 13 (Rotation/Rolling)

[SignaturePF]

Homework Statement

https://aapt.org/physicsteam/2012/upload/exam1-2011-1-3-answers_1.pdf
Number 13

Homework Equations

Rolling objects: E = 1/2mv^2 + 1/2Iw^2
I_cylinder = 1/2mR^2
Torque = rf(perp)
Torque = Iω = rfsinθ

The Attempt at a Solution

No idea how to proceed. Most likely starting with the torque equation.
T = rF = .02F

 

[Yenaled]

Think about what happens in non-rotational dynamics:
In order for the apparatus to not move horizontally, the friction force must equal the horizontal component of the force (Fcosθ) caused by the pulling of the thread.

Now, think in terms of rotational dynamics:
The torque produced by the pulling of the thread must equal the torque produced by the friction in order for the apparatus to not rotate.

What do you get from that?

After you have solved this problem, here's something else for you to ponder: What happens when θ is less than 60 degrees? Try it yourself with an actual yo-yo!

 

[SignaturePF]

Ok here we go
1) Tcosθ = Ff
2) Torque by friction = Torque by string
I'm confused about what the moment arm of friction is.
If we look at the 2nd image of the cylinder/disk system (cross-section)
I understand that the distance between point of rotation (the point touching ground) is √3 so the torque = √3Tsinθ
But what is friction's moment arm - wouldn't it have a zero moment arm if the point of rotation is the one touching the ground, since that is where friction is applied?

 

[SignaturePF]

I guess my question is: what is the point of rotation?

 

[Yenaled]

Ok here we go
But what is friction's moment arm - wouldn't it have a zero moment arm if the point of rotation is the one touching the ground, since that is where friction is applied?

Nope. The lever arm is the perpendicular distance from the axis of rotation to the point of force application.

I understand that the distance between point of rotation (the point touching ground) is √3 so the torque = √3Tsinθ

Once again, the axis of rotation is the point you should be concerned about, not the point touching the ground.

 

[SignaturePF]

Ok, so the center of rotation is the point in middle.
Net torque equation is:
(.02)Ff = .01Tsinθ
2Ff = Tsinθ
Combining this with the first equation:
Ff = Tcosθ gives:
2Tcosθ = Tsinθ
tanθ = 2
θ = 63.5 degrees -- this is not a valid answer.

 

[Yenaled]

You are almost there. ".01Tsinθ" is wrong. θ is not the angle between r and the force, T. What is the angle between r and T? Try drawing it out (look at the diagram on the F=MA test).

 

[SignaturePF]

Is the angle between r and the force 90 degrees?
Thus:
.02Ff = .01T
2Ff = T
2Tcostheta = T
2costheta = 1
costheta = 1/2
theta = 60 - This is the answer?

I think my understanding of torque/moment arms is poor. Could you explain how to find the moment arm given a non-perpendicular force in general please?

 

[Yenaled]

Good. r is perpendicular to the force.

Yeah, I think the main problem you were having was with moment arm angles.
Draw a line from the axis of rotation to the point of force application. The angle between that line and the line of force is the angle you should used.

Things will become much more clear when you study vectors and learn about cross products. rsin(angle) is perpendicular distance between the axis of rotation and the point of force application; it should always at a right angle to the force vector.

Here's a good explanation with diagrams:
http://hyperphysics.phy-astr.gsu.edu/hbase/torq2.html