Motion of a cylinder attatched to a spring

In summary, the problem involves determining the equation of motion and 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. The solution involves understanding that the spring extends by 2x, where x is the amount the center of the cylinder shifts, due to both the linear shift and the corresponding amount of arc length from the cylinder's rolling motion. This was initially misunderstood, but was clarified through a diagram and further explanation. The final understanding is that the spring stretches by 2x due to the combined effects of the cylinder's linear and rolling motion.
  • #1
polarcheese
2
0

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 [Broken]

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 [Broken]

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
 
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  • #2
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.
 
  • #3
ahhhhhhhhhhhhhhhhhh, amazing amazing.

Thank you so much, was getting myself really worked up about that.
 
  • #4
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!
 
  • #5
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
 

1. What is the motion of a cylinder attached to a spring?

The motion of a cylinder attached to a spring is typically described as simple harmonic motion, where the cylinder moves back and forth along a straight line due to the force of the spring.

2. What factors affect the motion of a cylinder attached to a spring?

The motion of a cylinder attached to a spring is affected by the spring constant, the mass of the cylinder, and the amplitude of the motion. Other external factors, such as friction and air resistance, can also impact the motion.

3. How is the period of the motion calculated for a cylinder attached to a spring?

The period of the motion can be calculated using the equation T = 2π√(m/k), where T is the period, m is the mass of the cylinder, and k is the spring constant.

4. Can the motion of a cylinder attached to a spring be damped?

Yes, the motion of a cylinder attached to a spring can be damped, meaning that the amplitude of the motion decreases over time due to the dissipation of energy. This can be caused by external factors such as friction or the internal friction of the spring itself.

5. How does the motion of a cylinder attached to a spring differ from a mass-spring system?

The motion of a cylinder attached to a spring is similar to that of a mass-spring system, but with the added factor of rotational motion of the cylinder. This can affect the calculation of the period and the amplitude of the motion.

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