Answer:
a. i. 3.43 m/s ii. 2.8 m/s
b. The thin-walled cylinder
Explanation:
a. Find translational speed of each cylinder upon reaching the bottom
The potential energy change of each mass = total kinetic energy gain = translational kinetic energy + rotational kinetic energy
So, mgh = 1/2mv² + 1/2Iω² where m = mass of object = 2.0 kg, g =acceleration due to gravity = 9.8 m/s², h = height of incline = 1.2 m, v = translational velocity of object, I = moment of inertia of object and ω = angular speed = v/r where r = radius of object.
i. translational speed of thin-walled cylinder upon reaching the bottom
So, For the thin-walled cylinder, I = mr², we find its translational velocity, v
So, mgh = 1/2mv² + 1/2Iω²
mgh = 1/2mv² + 1/2(mr²)(v/r)²
mgh = 1/2mv² + 1/2mv²
mgh = mv²
v² = gh
v = √gh
v = √(9.8 m/s² × 1.2 m)
v = √(11.76 m²/s²)
v = 3.43 m/s
ii. translational speed of solid cylinder upon reaching the bottom
So, For the solid cylinder, I = mr²/2, we find its translational velocity, v'
So, mgh = 1/2mv'² + 1/2Iω²
mgh = 1/2mv² + 1/2(mr²/2)(v'/r)²
mgh = 1/2mv'² + mv'²
mgh = 3mv'²/2
v'² = 2gh/3
v' = √(2gh/3)
v' = √(2 × 9.8 m/s² × 1.2 m/3)
v' = √(23.52 m²/s²/3)
v' = √(7.84 m²/s²)
v' = 2.8 m/s
b. Determine which cylinder has the greatest translational speed upon reaching the bottom.
Since v = 3.43 m/s > v'= 2.8 m/s,
the thin-walled cylinder has the greatest translational speed upon reaching the bottom.
