Auniform cylindrical hoop of mass 5.0 kg and radius 30 cm rolls without slipping, from rest, from the top of a 20 mhigh hill. ignoring losses due to friction and drag, the velocity of the cylinder at the bottom of the hill is (in m/s)

14 m/s

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

the answer is a i think

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..