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Step-by-Step Solution
Step 1: Identify the Rolling Objects
We have two cylinders of the same mass and same external diameter: one solid cylinder and one hollow cylinder. Both are released from rest at the same height on an inclined plane and roll without slipping.
Step 2: Write the General Formula for Rolling Time
The time taken by a rolling object to travel down an inclined plane of length $l$ at an angle $ \theta $ (without slipping) can be written as:
$ t = \sqrt{ \frac{2\,l \,\bigl[1 + \frac{K^2}{R^2}\bigr]}{g \,\sin\theta} } $
Here,
$g$ is the acceleration due to gravity.
$R$ is the radius of the cylinder.
$K^2 = \frac{I}{M}$, where $I$ is the moment of inertia and $M$ is the mass of the cylinder.
Step 3: Substitute Moment of Inertia for Each Cylinder
The radius of gyration squared, $K^2$, depends on the moment of inertia $I$ of the object:
Solid cylinder: $I_{\text{solid}} = \frac{1}{2} M R^2$, so
$K^2 = \frac{I}{M} = \frac{\frac{1}{2} M R^2}{M} = \frac{R^2}{2}.$
Hollow cylinder (thin-walled): $I_{\text{hollow}} = M R^2$, so
$K^2 = \frac{I}{M} = \frac{M R^2}{M} = R^2.$
Step 4: Compare the Times
For the solid cylinder, $1 + \frac{K^2}{R^2} = 1 + \frac{R^2/2}{R^2} = 1 + \frac{1}{2} = \frac{3}{2}.$
For the hollow cylinder, $1 + \frac{K^2}{R^2} = 1 + \frac{R^2}{R^2} = 1 + 1 = 2.$
Since the time $t$ is proportional to the square root of $\bigl[1 + \frac{K^2}{R^2}\bigr]$, we see that:
$ t_{\text{solid}} \propto \sqrt{\frac{3}{2}} $ and $ t_{\text{hollow}} \propto \sqrt{2}.
Numerically, $ \sqrt{\frac{3}{2}} < \sqrt{2}, $ so the time taken by the solid cylinder is less than that of the hollow cylinder.
Step 5: Conclusion
Because the solid cylinder has a smaller moment of inertia relative to its mass compared to the hollow cylinder, it accelerates faster and reaches the bottom of the inclined plane first. Thus, the solid cylinder arrives before the hollow cylinder.
