# How Do You Calculate the Kinetics of a Suspended, Unwrapping Cylinder?

• GleefulNihilism
In summary, the conversation discusses a problem involving a uniform cylinder suspended from a point with a string, and the resulting motion under the influence of gravity. The goal is to find the linear momentum, angular momentum, and kinetic energy of the cylinder. The conversation includes a problem statement, relevant equations, and an attempt at a solution. The final answers are given as follows: a) P = m*sprt((l'-r*phi')^2 + l^2*phi'^2), b) M = (3/2)*m*r*(l' - r*phi'), c) T = (3/4)*m*(l'-r*phi')^2 + (m/2)*l^2*phi'^
GleefulNihilism
Well, I had a couple problems on my final I was hoping to go over- hope nobody minds. Here's the first.

## Homework Statement

A uniform cylinder of radius r and mass m, wrapped around by an unstretchable and massless string, is suspended from a point. The cylinder comes down unwrapping the string and oscillating around the horizontal axis passing through the point under the action of gravity. Let l be the distance from the support point to the contact point of the string with the cylinder and phi be the angle the string forms with the vertical axis through the support point.
Find:
a.) The magnitude of the linear momentum of the cylinder.
b.) The angular momentum of the cylinder.
c.) The kinetic energy of the cylinder in terms of l, phi, dl/dt (written as l'), and d(phi)/dt (written as phi').

(Sorry, not sure how you do the fancier presentation codes.)

## Homework Equations

Linear Momentum
P = (m/2)*(x'^2+y'^2)

Angular Momentum
M = I*omega

Moment of Inertia for a Uniform Cylinder
I = (mr^2)/2

Kinetic Energy
T = (P^2)/2m + (M^2)/2mr^2

## The Attempt at a Solution

First, was part A.

P = (m/2)*(x'^2+y'^2)

The trick being finding x'^2 and y'^2. Preferably in terms of l, l', phi, and phi'.

Let's call the angle from the contact point on the cylinder to the center of the cylinder theta. Distance between these two points is always r- which is a constant in this problem. Woot!

So, x = l*sin(phi) + r*cos(theta) and y = l*cos(phi) + r*sin(theta).

Now, admittedly my weakest assumption, I assumed that the angle formed by the string to the contact point to the center of mass of the cylinder was usually around 90 degrees. Especially if the string wasn't close to being completely unwound. So, by method of similar triangles phi is about equal to theta.

Thus, roughly, x = l*sin(phi) + r*cos(phi) and y = l*cos(phi) + r*sin(phi).

Therefore x' = d(l*sin(phi) + r*cos(phi))/dt = l'*sin(phi) + l*phi'*cos(phi) - r*phi'*sin(phi)
Thus x'^2 = (l'*sin(phi) + l*phi'*cos(phi) - r*phi'*sin(phi))^2

And by similar arguments y'^2 = (l'*cos(phi) - l*phi'*sin(phi) + r*phi'*cos(phi))^2

So P = (m/2)*(x'^2 + y'^2)

P = (m/2)*((l'*sin(phi) + l*phi'*cos(phi) - r*phi'*sin(phi))^2 + (l'*cos(phi) - l*phi'*sin(phi) + r*phi'*cos(phi))^2)

Which expands into something that on the surface looks rather messy, but a few terms add out and other terms are simplified by the good old cos^2(theta) + sin^2(theta) = 1 identity.

So P = (m/2)*(l'^2 + l^2*phi'^2 + r^2*phi'^2 + 2*r*l'*phi'(cos^2(phi) - sin^2(phi)) - 4*r*l*phi'^2*sin(phi)*cos(phi))

B.

M=I*omega

For a uniform cylinder I = (mr^2)/2. Also known is omega = (v/r) = (P/mr)

Thus M = P*(r/2)

C.

T = (P^2)/2m + (M^2)/2mr^2

Which when you plug in P and M, do a little multiplication by constants, and you get.

T = (5m/32)*(l'^2 + l^2*phi'^2 + r^2*phi'^2 + 2*r*l'*phi'(cos^2(phi) - sin^2(phi)) - 4*r*l*phi'^2*sin(phi)*cos(phi))^2

No bites, eh?

Well, if it helps I don't need to be taken by the hand. I just want to see how the professor got these answers.

A) P = m*sprt((l'-r*phi')^2 + l^2*phi'^2)

B.) M = (3/2)*m*r*(l' - r*phi')

C.) T = (3/4)*m*(l'-r*phi')^2 + (m/2)*l^2*phi'^2

## 1. What is a Wrapped Cylinder Kinetics?

A Wrapped Cylinder Kinetics is a scientific term used to describe the movement and energy transfer of a cylindrical object that is wrapped or coated with a material. This material can affect the speed, direction, and other properties of the cylinder's motion.

## 2. How does a Wrapped Cylinder Kinetics work?

A Wrapped Cylinder Kinetics works by interacting with the surface of the cylinder and changing its characteristics. For example, a cylinder wrapped with a rough material may experience more resistance and slower motion compared to a smooth surface. The material can also affect the friction, heat, and other forces acting upon the cylinder.

## 3. What are the real-life applications of Wrapped Cylinder Kinetics?

Wrapped Cylinder Kinetics has various practical applications, such as in conveyor belts, engine components, and sports equipment. In conveyor belts, the material coating can reduce wear and tear, while in engine components, it can improve efficiency and reduce friction. In sports equipment, it can enhance performance and durability.

## 4. How is Wrapped Cylinder Kinetics different from regular cylinder motion?

The main difference between Wrapped Cylinder Kinetics and regular cylinder motion is the presence of an additional material that affects the cylinder's movement. In regular motion, the cylinder's properties remain the same, but in Wrapped Cylinder Kinetics, the properties can change due to the coating material.

## 5. What factors can affect the behavior of a Wrapped Cylinder Kinetics?

The behavior of a Wrapped Cylinder Kinetics can be affected by various factors, such as the type of coating material, the surface roughness of the cylinder, the speed of motion, and external forces like gravity and friction. Other factors like temperature, humidity, and pressure can also play a role in determining the behavior of a Wrapped Cylinder Kinetics.

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