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

• SignaturePF
In summary, the conversation discusses the concept of torque and how it applies to rotational dynamics. The problem of finding the angle at which a yo-yo will remain stationary when pulled by a string is discussed, and the correct answer is determined by considering the perpendicular distance between the axis of rotation and the point of force application. The importance of understanding moment arms and their angles is emphasized, and a resource with diagrams is provided for further explanation.
SignaturePF

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

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!

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?

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

SignaturePF said:
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.

SignaturePF said:
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.

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.

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).

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?

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

## 1. What is the equation F = MA used for?

The equation F = MA is used to calculate the force (F) required to accelerate an object with a certain mass (M) at a certain rate of acceleration (A). It is one of the fundamental equations in the study of mechanics and is commonly known as Newton's second law of motion.

## 2. What does the "R" in F = MA Exam 2011 # 13 (Rotation/Rolling) stand for?

In this context, the "R" represents rotation or rolling. This means that the question is likely related to a scenario where an object is rotating or rolling, and the equation F = MA will be used to solve for the force involved in that motion.

## 3. How does the F = MA equation apply to rotation/rolling?

The F = MA equation can be applied to rotation/rolling by using the rotational equivalent of force, known as torque (T). The equation for torque is T = Iα, where I is the moment of inertia of the object and α is the angular acceleration. By substituting this into the original equation, we get T = Iα = MR²α, which is the rotational equivalent of F = MA.

## 4. Can F = MA be used for both linear and rotational motion?

Yes, F = MA can be used for both linear and rotational motion. In linear motion, the equation is typically written as F = MA, where M is the mass and A is the linear acceleration. In rotational motion, the equation is written as T = Iα, where T is the torque, I is the moment of inertia, and α is the angular acceleration. However, the two equations are equivalent and can be used interchangeably.

## 5. What are some real-life applications of F = MA in rotation/rolling scenarios?

F = MA has many real-life applications in rotation/rolling scenarios. Some examples include calculating the force required to rotate a bicycle wheel, the force needed to roll a ball up a ramp, or the force required to spin a top. This equation is also used in designing machines, such as gears and pulleys, which involve rotational motion.

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