Question

Two uniform solid spheres, each with mass 0.800 kg and radius 8.00×10⁻² m, 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 157 N/m 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.

Answer 1

Relevant equations τ=Iα

F = ma

Fx=-kx

M = mass total

ω=sqrt(k/m)

The attempt at a solution

EF = Fstatic-Fx = Ma

Fstatic-kx = Ma
I = 2/5MR^2+ MR^2

I = 7/5MR^2 (Moment of inertia around contact with Ground because it rolls without sliding.

-Fstatic= 7/5 Ma

With -Fstatic 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 = ω2x

so ω = sqrt(5k/12M)

Finding period T

T = 2pi/ω

or 2pi * sqrt(12M/5k)

or 2*3.14*sqrt(12*0.800/5*157)

Putting in m and k I found 0.694 s ans