Motion of a cylinder attatched to a spring

[polarcheese]

Homework Statement

Determine the equation of motion and the natural frequency of vibration for A uniform cylinder of mass M, rolling without slipping on a horizontal plane and constrained by a spring of stiffness K, as shown.

http://img638.imageshack.us/img638/613/cylinder.png

2. The attempt at a solution

So I am now in full on revision mode, and I have the worked solutions to the problem:

http://img571.imageshack.us/img571/9987/cylinder1.png

Although I can follow most of the solution, I don't understand why the spring extends by 2x if the centre moves by x?

Thanks very much

 

[gneill]

However much x the center of the cylinder shifts, since it rolls without slipping it must turn through an angle so that it rolls along an equal length x of its circumference. Where the spring attaches experiences both the linear shift and the corresponding amount of arc length. If you draw a diagram showing the profile of the cylinder in both positions, you'll see that the result is a total length of 2x.

 

[polarcheese]

ahhhhhhhhhhhhhhhhhh, amazing amazing.

Thank you so much, was getting myself really worked up about that.

 

[alice.w]

I'm really sorry to bring this up, but I have found out where I have been going wrong in this problem - and it is the 2x bit, but I still don't understand why the spring stretches by 2x?
So the spring moves horizontally x, so therefore the spring stretches by x, but since the cylinder is rolling... ah I think I understand having posted this, sorry!

 

[rwisz]

for your help, it's much appreciated.

I can provide a response to the content by explaining the physics behind the motion of a cylinder attached to a spring.

Firstly, let's consider the set-up of the system. We have a uniform cylinder of mass M, rolling without slipping on a horizontal plane. This means that the cylinder has both translational and rotational motion. The cylinder is also constrained by a spring of stiffness K, which means that there is a force acting on the cylinder in the opposite direction to its displacement.

Now, let's look at the equation of motion for this system. The force acting on the cylinder is the sum of the force from the spring and the force from the cylinder's motion. This can be written as:

F = -Kx + ma

Where F is the total force, K is the stiffness of the spring, x is the displacement of the cylinder from its equilibrium position, m is the mass of the cylinder, and a is the acceleration of the cylinder.

Since the cylinder is rolling without slipping, we can use the relationship between linear and angular motion:

v = ωr

Where v is the linear velocity of the cylinder, ω is the angular velocity, and r is the radius of the cylinder. We can also relate the acceleration of the cylinder to its angular acceleration:

a = αr

Where α is the angular acceleration.

Now, if we consider the torque acting on the cylinder, we can write:

τ = Iα

Where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

For a uniform cylinder, the moment of inertia is given by:

I = ½MR^2

Where R is the radius of the cylinder.

Combining these equations, we can write the equation of motion for the cylinder as:

-Kx + ½MR^2α = Ma

Now, let's look at the natural frequency of vibration for this system. The natural frequency is the frequency at which the system will vibrate without any external forces acting on it. In this case, it is the frequency at which the cylinder will oscillate back and forth due to the force from the spring.

The natural frequency can be calculated using the equation:

ω = √(K/M)

Where ω is the natural frequency, K is the stiffness of the spring, and M is the mass of the cylinder.

To answer your question about why the spring extends by