- #1

OlicityFangirl

## Homework Statement

A solid cylinder of mass, M, is connected to two springs of total stiffness, k. The springs are connected tangentially (on top) to the cylinder. The other ends of the springs are attached to walls. What is the period of oscillation of the cylinder assuming that it does not slide on the floor?

## Homework Equations

τ=Iα=F

_{tan}R

F

_{spring}=-kx

a=Rα

T= 2π/ω

## The Attempt at a Solution

So the springs are going to provide a tangential force on the cylinder, which will cause it to rotate. I'm struggling a little bit with the concept of two springs, but I did some googling and it just seems like they act as one spring with a combined spring constant, which is known in this problem. So force of the springs is going to apply a torque:

-kxR=τ

Torque can also be expressed in terms of moment of inertia and angular acceleration.

Iα=τ

I can set these two guys equal to each other:

Iα=-kxR

Moving them all to one side to get a first-order differential equation:

Iα+kRx

Angular acceleration is a=Rα, so I can substitute and use x'' for a:

Ix''/R + kRx = 0

Then to have x'' on its own, I multiply by R and divide by I:

x'' + [kR

^{2}/I]*x = 0

Using the moment of inertia of a cylinder: I=0.5MR

^{2}:

x'' + [2kR/MR

^{2}]x=x''+ [2k/M]x=0

This would make my angular frequency, ω, (2k/M)

^{0.5}, and I can use T= 2π/ω to get

T=2π/(2k/M)

^{2}

I'm concerned that the differential equation should be different because there are two springs involved, but I'm not really sure how. The way I solved the problem resembles how I would solve an oscillation problem with one spring too much :P