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A solid brass cylinder and a solid wooden cylinder have the same radius and mass, the wooden cylinder being longer. You release them together at the top of an incline. Which will beat the other to the bottom? Suppose that the cylinders are now made to be the same length (and radius) and that the masses are made to be equal by boring a hole along the axis of the brass cylinder. Which cylinder will win the race now? Explain your answers. Assume that the cylinders roll without slipping.

6 years ago

assumed axis is

It is important to note that the rotational inertia of the cylinder about the assumed axis is independent of its length L .

If the solid brass cylinder has the same mass and radius as the wooden cylinder, then both the cylinders will reach the bottom of inclined plane together because, the rotational inertia of the cylinders will be same, which is

Even though the net mass of both the cylinders after boring a hole in brass cylinder is same, the rotational inertia of the brass cylinder will be larger. This is because the rotational inertia of the brass cylinder is mainly contributed from the particles away from the axis of rotation.

One can treat the brass cylinder as an annular cylinder with radius R

Thus, this time the wooden cylinder will win the race.

6 years ago

In the first both will reach at same time since they have same moment of inertia.

In the second case brass rod will be the first to reach bottom.

Acceleration on incline plane can be calculated using this eqn; a = (gsinθ)/(1+(I/MR^2))

I= moment of inertia

In first case both have same value of I.

But in second case I=(M(R^2 – r^2))/2 for brass cylinder.

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