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Simple Harmonic Motion (Mass on a Spring)

  • Thread starter seichan
  • Start date
32
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[SOLVED] Simple Harmonic Motion (Mass on a Spring)

1. 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


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

3. 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.
 

Answers and Replies

49
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You actually don't need radius of cylinder. At the begining 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.
 
32
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Thank you very much!
 
57
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Re: [SOLVED] Simple Harmonic Motion (Mass on a Spring)

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

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