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 they 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

Answer:

a. They will be tie

b. Win the wood cylinder

Explanation:

a.

The both cylinders will reach the bottom at the same time notice the relation in the equation in indepent of the length and both have the same radius and the same rotational inertia.

$I=\frac{1}{2}*m*r^2$

$a=\frac{g*sin(\beta)}{1+I_{com}/m*r^2}$

So both will be tie

b.

$a_{brass}=\frac{g*sin(\beta)}{1+I_{brass}/m*r^2}=a_{wood}=\frac{g*sin(\beta)}{1+I_{wood}/m*r^2}$

The acceleration of the wood cylinder is larger than the acceleration of the brass cylinder so the cylinder of wood will reach the bottom first

$a_{brass}<a_{wood}$

So the wood win the race