Simple Harmonic Motion (Mass on a Spring)

In summary, the problem involves a solid cylinder attached to a horizontal spring, with a force constant of 346 N/m. The system is released from rest at a position where the spring is stretched by 17.0 cm. To find the translational kinetic energy when it passes through equilibrium position, the formula is 0.5mv^2. For rotational kinetic energy, the formula is 0.5Iw^2, where the angular velocity is found using sqrt(k/m). The radius of the cylinder is not needed in the calculations.
  • #1
seichan
32
0
[SOLVED] Simple Harmonic Motion (Mass on a Spring)

Homework Statement


A solid cylinder of mass M= 10.8 kg is attached to a horizontal massless spring so that it can roll without slipping along a horizontal surface, as shown in the Figure. The force constant of the spring is k= 346 N/m. The system is released from rest at a position in which the spring is stretched by a distance x= 17.0 cm. What is the translational kinetic energy of the cylinder when it passes through the equilibrium position? What about the rotational KE?

http://i3.photobucket.com/albums/y65/amenochikara/prob02a.gif


Homework Equations


Translational Energy-.5mv^2
Rotational Energy- .5Iw^2
Angular Velocity (spring)- sqrt(k/m)

The Attempt at a Solution



Alright... I only have this problem left and it's frustrating me. For the first part, I used translational Energy for a spring (.5kx^2), but I am not taking the mass into account. All the example problems we were given have amplitudes and other useful things to use to compute velocity. So, I moved onto what I thought would be the easier one, rotational. For this, I tried using .5(r^2)(k) when I realized, much to my dismay, that I had no radius for the cylinder. Any alternative ways I am not thinking of? Thanks so much.
 
Physics news on Phys.org
  • #2
You actually don't need radius of cylinder. At the beginning you have just energy of spring
0.5kx^2 , that equals sum of rotational and translational energy in equilibrium position. The you should know that

[tex] \omega =v/r ~,~E_t=0.5mv^2~,~E_r=0.5I~\omega^2 [/tex]

Now since I for cylinder is 0.5mr^2 , r falls out. You have only equation for v then.
 
  • #3
Thank you very much!
 
  • #4


Can someone please explain to me where the formulas are coming from for translational energy, etc?
 
  • #5




Simple harmonic motion is a type of periodic motion in which the restoring force is directly proportional to the displacement from equilibrium and is directed towards the equilibrium point. In this case, the mass on a spring system is a classic example of simple harmonic motion.

In order to solve this problem, we first need to find the equilibrium position of the cylinder. This can be done by setting the net force on the cylinder to be zero, which gives us the equation kx = mg, where k is the force constant of the spring, x is the displacement from equilibrium, and mg is the weight of the cylinder. Solving for x, we get x = mg/k = (10.8 kg)(9.8 m/s^2)/(346 N/m) = 0.312 m.

Next, we can use the given displacement of 17.0 cm to find the total displacement from equilibrium, which is x + 0.17 m. This will be the amplitude of the motion.

To find the translational kinetic energy of the cylinder at the equilibrium position, we can use the equation .5mv^2. However, since the cylinder is rolling without slipping, we need to use the equation for the translational kinetic energy of a rolling object, which is .5mv^2 + .5Iw^2, where I is the moment of inertia and w is the angular velocity. In this case, the moment of inertia for a solid cylinder is 1/2MR^2, where R is the radius. Since we do not have the radius, we can use the relationship between the angular velocity and the spring constant, which is w = sqrt(k/m). Substituting this into the equation, we get .5mv^2 + .5(1/2MR^2)(k/m) = .5mv^2 + .5(k/m)(v/R)^2. Now, we can use the fact that at the equilibrium position, the velocity of the cylinder is zero, so the translational kinetic energy is also zero.

For the rotational kinetic energy at the equilibrium position, we can use the equation .5Iw^2, where I is the moment of inertia and w is the angular velocity. Again, since we do not have the radius, we can use the relationship between the angular velocity and the spring constant, which is w = sqrt(k/m). Substituting this into the equation, we get .5I(k/m
 

1. What is Simple Harmonic Motion (SHM)?

Simple Harmonic Motion is a type of periodic motion where an object oscillates back and forth in a straight line with a constant amplitude and period. It follows a sinusoidal pattern and is characterized by the restoring force being directly proportional to the displacement from the equilibrium position.

2. What is a mass on a spring system?

A mass on a spring system is a simple physical system where a mass is attached to one end of a spring and the other end is fixed. The system exhibits SHM as the mass moves back and forth along the spring due to the restoring force of the spring.

3. What factors affect the period of a mass on a spring system?

The period of a mass on a spring system is affected by the mass of the object, the spring constant, and the amplitude of the oscillation. The period is longer for a larger mass, a stiffer spring, and a larger amplitude.

4. How is the frequency of a mass on a spring system calculated?

The frequency of a mass on a spring system is calculated by dividing the inverse of the period (T) by 2π. This can also be represented as ω = 2π/T, where ω is the angular frequency.

5. What is the relationship between potential and kinetic energy in SHM?

In SHM, there is a continuous exchange between potential energy and kinetic energy. When the mass is at the equilibrium position, it has maximum potential energy and no kinetic energy. As it moves towards the maximum displacement, potential energy decreases and kinetic energy increases. At the maximum displacement, potential energy is at its minimum and kinetic energy is at its maximum. The process then repeats as the mass moves back towards equilibrium.

Similar threads

  • Introductory Physics Homework Help
Replies
16
Views
333
  • Introductory Physics Homework Help
2
Replies
51
Views
2K
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
803
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
274
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
2K
Replies
13
Views
262
  • Introductory Physics Homework Help
Replies
29
Views
822
Back
Top