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

Answer:

the sphere

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

$F = \dfrac{I \alpha }{R}$

where

$\alpha = \dfrac{a}{R}$  in a rolling motion

Then;

$F = \dfrac{I a }{R^2}$

∴

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

$\mathtt{mg sin \ \theta- \dfrac{Ia}{R^2} = ma}$

making a the subject of the formula, we have:

$a = (\dfrac{mg \ sin \theta}{m + \dfrac{I}{R^2}})$

Similarly;

I = mk²  in which k is the radius of gyration

∴

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

$a = (\dfrac{mg \ sin \theta}{m + \dfrac{mk^2}{R^2}})$

where;

the uniform disk $\dfrac{k^2}{R^2 }= \dfrac{1}{2}$  

the uniform  sphere $\dfrac{k^2}{R^2 }= \dfrac{2}{5}$

∴

$a = \dfrac{2}{3} \ g sin \theta \ for \ the \ uniform \ disk$

$a = \dfrac{5}{7} \ g sin \theta \ for \ the \ uniform \ sphere$

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