Question

A small uniform disk and a small uniform sphere are released simultaneously at the top of a high inclined plane, and they roll down without slipping. Which one will reach the bottom first?

Answer 1

the sphere

Step-by-step explanation:

From the given information,

A free flow body diagrammatic expression for the small uniform disk and a small uniform sphere which are released simultaneously at the top of a high inclined plane can be seen in the image attached below.

From the diagram;

The Normal force mgsinθ - Friction force F = mass m × acceleration a

Meanwhile; the frictional force

`F = (I \alpha )/(R)`

where

`\alpha = (a)/(R)` in a rolling motion

Then;

`F = (I a )/(R^2)`

∴

The Normal force mgsinθ - F = m × a can be re-written as:

`\mathtt{mg sin \ \theta- (Ia)/(R^2) = ma}`

making a the subject of the formula, we have:

`a = ((mg \ sin \theta)/(m + (I)/(R^2)))`

Similarly;

I = mk² in which k is the radius of gyration

∴

replacing I = mk² into the above equation , we have:

`a = ((mg \ sin \theta)/(m + (mk^2)/(R^2)))`

where;

the uniform disk
`(k^2)/(R^2 )= (1)/(2)`

the uniform sphere
`(k^2)/(R^2 )= (2)/(5)`

∴

`a = (2)/(3) \ g sin \theta \ for \ the \ uniform \ disk`

`a = (5)/(7) \ g sin \theta \ for \ the \ uniform \ sphere`

We can now see that the uniform sphere is greater than the disk as such the sphere will reach the bottom first.