Cylinder+String Rolling w/o slipping

[Imuell1]

Homework Statement

A uniform cylinder of mass M and radius R has a string wrapped around it. The string is held fixed, and the cylinder falls vertically. (a) Show that the acceleration of the cylinder is downward with a magnitude a=2g/3. (b) Find the tension in the string.

Homework Equations

I=1/2MR²

The Attempt at a Solution

Not sure how to start, does it involve the conservation of mechanical energy?

 

[CaptainEvil]

Try Newton's methods.

Draw a freebody diagram for the cylinder, include all the forces acting on it and use Newton's second law, along with torque = Ialpha, to solve for your unknowns

 

[nlightner]

As a scientist, the first step in solving this problem would be to identify the relevant equations and principles that apply to this situation. In this case, the conservation of energy and the concept of rolling without slipping are key factors.

First, we can use the fact that the cylinder is falling vertically to determine its acceleration. Since there are no external forces acting on the cylinder, we can use the conservation of energy to equate the potential energy at the top (when the cylinder is released) to the kinetic energy at the bottom (when the cylinder has reached its maximum velocity). This can be written as:

mgh = 1/2mv^2

Where m is the mass of the cylinder, g is the acceleration due to gravity, h is the height of the cylinder, and v is the velocity of the cylinder. We can rearrange this equation to solve for the acceleration:

a = v^2/2h = 2gh/2h = g

This shows that the acceleration of the cylinder is indeed equal to the acceleration due to gravity, and it is in the downward direction.

Next, we can use the concept of rolling without slipping to determine the tension in the string. This means that the point of contact between the cylinder and the surface it is rolling on has zero velocity. In this case, the surface is the string, so the point of contact is where the string is held fixed. This allows us to write the following equation:

v = ωR

Where ω is the angular velocity of the cylinder and R is the radius of the cylinder. We can also write the equation for the acceleration of the point of contact as:

a = αR

Where α is the angular acceleration of the cylinder. Since the cylinder is falling vertically, the angular acceleration can be related to the linear acceleration using the following equation:

a = αR = rα = g

Where r is the radius of the cylinder. We can now substitute this into our first equation to solve for the angular velocity:

v = ωR = gR/R = g

Finally, we can use this value for the angular velocity to determine the tension in the string. We can write the equation for the tension as:

T = m(a + g) = mg + mrα = mg + mgr/R = 2mg/3

Where m is the mass of the cylinder and r is the radius of the cylinder. This shows that the tension in the string is equal to