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
[tex]F = \dfrac{I \alpha }{R}[/tex]
where
[tex]\alpha = \dfrac{a}{R}[/tex] in a rolling motion
Then;
[tex]F = \dfrac{I a }{R^2}[/tex]
∴
The Normal force mgsinθ - F = m × a can be re-written as:
[tex]\mathtt{mg sin \ \theta- \dfrac{Ia}{R^2} = ma}[/tex]
making a the subject of the formula, we have:
[tex]a = (\dfrac{mg \ sin \theta}{m + \dfrac{I}{R^2}})[/tex]
Similarly;
I = mk² in which k is the radius of gyration
∴
replacing I = mk² into the above equation , we have:
[tex]a = (\dfrac{mg \ sin \theta}{m + \dfrac{mk^2}{R^2}})[/tex]
where;
the uniform disk [tex]\dfrac{k^2}{R^2 }= \dfrac{1}{2}[/tex]
the uniform sphere [tex]\dfrac{k^2}{R^2 }= \dfrac{2}{5}[/tex]
∴
[tex]a = \dfrac{2}{3} \ g sin \theta \ for \ the \ uniform \ disk[/tex]
[tex]a = \dfrac{5}{7} \ g sin \theta \ for \ the \ uniform \ sphere[/tex]
We can now see that the uniform sphere is greater than the disk as such the sphere will reach the bottom first.
