Answer: 14 m/s
Step-by-step explanation:
If we ignore losses due to friction and drag, the change in gravitational potential energy must be equal to the change of kinetic energy of the object.
As the object is a rigid body, it has a traslational kinetic energy, and a rotational one.
So, we can write the following expression:
ΔK = Kf -Ki
If the body starts from rest, this means that Ki = 0, so:
ΔK = 1/2 mv² + 1/2 I ω² = ΔU = m.g.h
Now, as the cylinder rolls without slipping, this means, that at any time, there exists a relationship between v and ω (linear velocity and angular velocity) which is the same than in a uniform circular motion, i.e.,
ω = v/r
Replacing in the expression of ΔK:
ΔK = 1/2 mv² + 1/2 I (v/r)²
Now, as in a cylindrical hoop all the mass is concentrated on the outer ring, the roatational inertia is directly I = mr².
Replacing in ΔK:
ΔK = 1/2 mv² + 1/2 mr².(v/r)² = 2 (1/2 mv²) = mv²
As ΔK = ΔU, we can write:
mv² = m.g.h⇒ v=√g.h = √9.8 m/s².20 m = 14 m/s