General Formula for Acceleration of a Disk Rolling Down an Incline

In summary, the conversation discusses a student's experiment involving modeling the acceleration of a disk rolling down an incline with different hole sizes in the center. The student is aware of the moment of inertia, net force vector, and torque force, but is unsure of the value for angular acceleration. The final formula given by the teacher involves gsin(θ)/ (1/2(a/b)^2 + 3/2) and the student is struggling to find a relation between angular and linear acceleration. The conversation concludes by mentioning the importance of the radius of the disk in this experiment.
  • #1
theboomking
1
0
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?
 
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  • #2
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.
 

1. What is the general formula for acceleration of a disk rolling down an incline?

The general formula for acceleration of a disk rolling down an incline is a = g*sin(θ), where a is the acceleration, g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the incline.

2. How does the angle of the incline affect the acceleration of a rolling disk?

The angle of the incline directly affects the acceleration of a rolling disk. As the angle increases, the acceleration also increases. This is because the component of the force of gravity acting down the incline (mg*sin(θ)) increases with a larger angle, resulting in a larger acceleration.

3. Is the acceleration of a rolling disk affected by the mass of the disk?

The acceleration of a rolling disk is not affected by the mass of the disk. This is because the mass cancels out in the general formula for acceleration, leaving only the angle of the incline and the acceleration due to gravity as factors.

4. How does the radius of the disk affect its acceleration down an incline?

The radius of the disk does not directly affect its acceleration down an incline. However, a larger radius may result in a larger moment of inertia, which would require a larger torque to overcome and therefore decrease the acceleration.

5. Can this formula be applied to objects other than disks?

Yes, this formula can be applied to objects other than disks, as long as they are rolling without slipping down an incline. This includes objects such as spheres, cylinders, and other rolling objects with rotational inertia. The key is that the object must be rolling without slipping, otherwise the formula will not be accurate.

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