Basics Of Rolling

For any body of mass $M$ and radius $R$ , we can write it's moment of inertia as $M {K}^{2}$ , where $K$ is the radius of gyration for that body. This argument is enough for the bodies we will be considering.

Now, for any body, the force that accelerates it is the component of it's weight along the inclined, which can be written as $Mg \sin{\theta}$ where $\theta$ is the angle, the incline makes with the horizontal. Since, the body rolls perfectly, hence, there must be an opposite frictional force. Let that force be $f$ . This frictional force decelerates the body linearly, but provides it an accelerating torque.

Writing down the equations of motion along the plane:

$Mg \sin{\theta} - f = Ma$

where $a$ is the linear acceleration of the body.

Secondly, for rotational motion:

$f \times R = M{K}^{2} \alpha$

where $\alpha$ is the angular acceleration of the body.

Now, using the relation $\alpha = \frac {a}{R}$ in the second equation, we get

$f = Ma \frac{{K}^{2}} {{R}^{2}}$

Using this in the first equation, we can easily get:

$a = \frac {g\sin{\theta}} {1+\frac {{K}^{2}} {{R}^{2}}}$

Hence, the body with more radius of gyration will have a lower acceleration, and hence will take more time to roll down the incline.

Comparing the radii of gyration of all the given bodies, we can easily make out that

${K}_{Solid Sphere} = \frac {2}{5} < {K}_{Hollow Sphere} = \frac {2}{3} < {K}_{Solid Cylinder} = \frac {1}{2} < {K}_{Hollow Cylinder} = 1$

And so,

${time}_{Solid Sphere} < {time}_{Hollow Sphere} < {time}_{Solid Cylinder} < {time}_{Hollow Cylinder}$