Question

An empty cylindrical barrel is open at one end and rolls without slipping straight down a hill. The barrel has a mass of 19.0 kg
a radius of 0.400 m, and a length of 0.600 m. The mass of the end of the barrel equals a third of the mass of its side, and the
thickness of the barrel is negligible.
The acceleration due to gravity is g = 9.80 m/s².
What is the translational speed up of the barrel at the bottom
of the hill if released from rest 27.0 m above the bottom?
Uf => m/s

Answer 1

Answer:

16.8 m/s

Explanation:

Energy is conserved.

Initial potential energy = final kinetic energy + rotational energy

PE = KE + RE

mgh = ½ mv² + ½ Iω²

The side of the barrel is a hollow cylinder, and the end of the barrel is a solid disk.

For a hollow cylinder, I = mr².

For a solid disk, I = ½ mr².

The mass of the end is one third the mass of the side, or one fourth the total mass. The mass of the side is three-fourths the total mass. Therefore, the total moment of inertia of the barrel is:

I = (¾ m) r² + ½ (¼ m) r²

I = ⅞ mr²

For rolling without slipping, ω = v / r.

Plugging in:

mgh = ½ mv² + ½ (⅞ mr²) (v/r)²

mgh = ½ mv² + ⁷/₁₆ mv²

mgh = ¹⁵/₁₆ mv²

gh = ¹⁵/₁₆ v²

v² = ¹⁶/₁₅ gh

Given g = 9.80 m/s² and h = 27.0 m:

v² = ¹⁶/₁₅ (9.80 m/s²) (27.0 m)

v = 16.8 m/s