Simple Harmonic Motion (Mass on a Spring)

AI Thread Summary
The discussion centers on solving a physics problem involving a solid cylinder attached to a spring, focusing on calculating its translational and rotational kinetic energy as it passes through the equilibrium position. The user initially struggles with incorporating mass into the energy equations and is unsure how to proceed without the cylinder's radius. It is clarified that the initial potential energy stored in the spring can be equated to the sum of the translational and rotational kinetic energies at equilibrium. The user is reminded that the radius cancels out in the equations, simplifying the calculations. Understanding the derivation of the formulas for translational and rotational energy is also requested.
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[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.
 
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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

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

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


Can someone please explain to me where the formulas are coming from for translational energy, etc?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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