Rolling Sphere Oscillating on Spring

[timmastny]

Homework Statement

Two uniform solid spheres, each with mass 0.862 and radius 8.00×10−2 , are connected by a short, light rod that is along a diameter of each sphere and are at rest on a horizontal tabletop. A spring with force constant 164 has one end attached to the wall and the other end attached to a frictionless ring that passes over the rod at the center of mass of the spheres, which is midway between the centers of the two spheres. The spheres are each pulled the same distance from the wall, stretching the spring, and released. There is sufficient friction between the tabletop and the spheres for the spheres to roll without slipping as they move back and forth on the end of the spring.

Assume that the motion of the center of mass of the spheres is simple harmonic. Calculate its period.

Homework Equations

[itex]\tau[/itex]=I[itex]\alpha[/itex]
F = ma
F[itex]_{x}[/itex]=-kx

M = mass total

[itex]\omega[/itex]=sqrt(k/m)

The Attempt at a Solution

[itex]E[/itex]F = F[itex]_{static}[/itex]-F[itex]_{x}[/itex] = Ma

F[itex]_{static}[/itex]-kx = Ma

I = 2/5MR[itex]^{2}[/itex] + MR [itex]^{2}[/itex]
I = 7/5MR[itex]^{2}[/itex] (Moment of inertia around contact with Ground because it rolls without sliding.

[itex]\tau[/itex][itex]_{tot}[/itex]=-F[itex]_{static}[/itex]R = 7/5MR[itex]^{2}[/itex](a/R)

-F[itex]_{static}[/itex]= 7/5 Ma

With -F[itex]_{static}[/itex] in hand, I placed it back into the translation force equation

-7/5Ma -kx = Ma
and
-kx = 12/5 Ma
Solving for a

a = -(5k/12M)x

now a = [itex]\omega[/itex][itex]^{2}[/itex]x so

[itex]\omega[/itex] = sqrt(5k/12M)

Finding period T

T = 2pi/[itex]\omega[/itex]
or 2pi * sqrt(12M/5k)

Putting in m and k I found .706 s

----
Since the problem says their are two spheres touching the ground, would I have to alter my answer? I don't know how rolling without sliding works with two bodies on an axle. Would the moment of inertia be doubled? Thanks for the help

This is my first time, sorry if my formatting is off. I attempted to make it as clear as possible.

 

[timmastny]

UPDATE: I couldn't figure out what was wrong with my previous answer, but I found something I did wrong in my calculators. I was using big M to denote total mass.

However in my calculations I was multiplying by .862

Revised with total mass 1.724 and plugging into 2*pi*sqrt(12M/5K) I found

T = .998

EDIT: Still incorrect

 

[timmastny]

UPDATE: Rereading the section on Rolling without Slipping, I'm not sure I'm supposed to set
F[itex]_{static}[/itex]R = I[itex]\alpha[/itex]

where I follows the parallel axis theorem.

So instead of I = 2/5MR[itex]^{2}[/itex]+MR[itex]^{2}[/itex] I think it should just be equal to I = 2/5MR[itex]^{2}[/itex]

That said, I have tried this solution but I was still caring the mistake I found in my last updated. With Updated values for I, I find

[itex]\omega[/itex]=sqrt(5k/7M)

and

T = 2*pi*sqrt(7M/5K)

Pulling in TOTAL mass of spheres and constant K, I find

T = .762

--- does anyone think this looks correct? I'm still very unsure because there are technically two spheres rolling on the ground, and I am at a complete loss at how to model that.

EDIT: I was right.