- #1

RiotRick

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## Homework Statement

A solid cylinder rolls down an inclined plane because of stiction. How does angular velocity change over time?

Given:

##m,R,g,\alpha,\mu## where ##\mu## is the friction coefficient

## Homework Equations

[/B]

##J_{cm} = 1/2 m*r^2## moment of inertia

m = mass of the cylinder

Parallel Axis theorem: ##J=J_{cm}+mr^2##

## M = I*d\omega/dt ## = Torque

## The Attempt at a Solution

First I thought I take the torque at the center of mass pulling it down

##R*mg*sin(\alpha)=d\omega/dt*(J_{cm}+mr^2)##

which gives me as an end result:

## d\omega/dt = \frac{2*g*sin(\alpha)}{3*r} ##

= ## \omega = \frac{2*g*sin(\alpha)}{3*r}*t ##

Then I thought it would have been simpler if I looked at the point of friction:

## M/J = r \times F / J = r*F_{friction}/J = \mu*r*mg*cos(\alpha)/J=d\omega/dt ##

=## \frac{2*\mu*g*cos(\alpha)}{3r}=d\omega/dt ##

=## \frac{2*\mu*g*cos(\alpha)}{3r}*t=\omega ##

Are both correct, one of them or both garbage?