Question

A solid brass cylinder and a solid wood cylinder have the same radius and mass (the wood cylinder is longer). Released together from rest, they roll down an incline. (a) Which cylinder reaches the bottom first, or do the tie? (b) The wood cylinder is then shortened to match the length of the brass cylinder, and the brass cylinder is drilled out along its long (central) axis to match the mass of the wood cylinder. Which cylinder now wins the race, or do they tie?

Answer 1

The general formula for the acceleration for a mass m along a ramp of inclination is,
a = g sin / [1 + I/mr²]
Where I is the moment of inertia and r is the radius of mass.

a)
In the first case, both the brass cylinder and wood cylinder have same mass and same radius.
So both have same moment of inertia, I = 1/2 mr²
a = g sin / [1 + (1/2 mr²) / (mr²)]
= g sin / 1.5
Since acceleration is same for both the masses, they take same time to reach the bottom. So it is a tie.

b)
Consider that both have mass m.
Moment of inertia of the wood cylinder, Ic = mr²
Acceleration of the wood cylinder, a = g sin / [1 + (1/2 mr²) / (mr²)]
= g sin / 1.5

The brass cylinder becomes a shell, its new moment of inertia, Is = mr²
Acceleration, a = = g sin / [1 + (mr²) / (mr²)]
= g sin / 2

So the wood has more acceleration and it will reach the bottom first.