General Formula for Acceleration of a Disk Rolling Down an Incline

1. Jun 3, 2014

theboomking

Hi everyone,

For an experiment I am writing up I have to model the acceleration of a disk rolling down an incline (at an angle of θ to the ground) with various hole sizes in the center.

A am aware that the moment of inertia for this object is given by 1/2m(a^2 + b^2) and that the net force vector down the incline will be given by mgsin(θ) - F (friction)= ma

I am also aware that the torque force will be equal to the moment of inertia times the angular acceleration, which will also be equal to the force of friction times b. However I am not very familiar with angular acceleration and cannot work out what the value should be. The final formula my teacher gave me looks like this

gsin(θ)/ (1/2(a/b)^2 + 3/2)

I've tried all substituting an expression for the force of friction into the net force value, but can't get it to work as I am not sure of the angular acceleration value. Can anybody help me?

2. Jun 3, 2014

dauto

You need a relation between angular acceleration and linear acceleration which depends on the radius of the disk. Assuming no slip condition,we have a=αR, where a is the linear acceleration, α is the angular acceleration, and R is the radius of the disk. Based on what you said it seems that R = b, but you should make sure that is indeed correct. It depends on the exact geometry of your disk and experimental set up.

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