g A 5.60-kilogram hoop starts from rest at a height 1.80 m above the base of an inclined plane and rolls down under the influence of gravity. What is the linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface? (Neglect friction.)

Linear speed = 4.2 m/s

Explanation:

Potential Energy at top = Kinetic Energy at bottom

Now, KE is the sum of linear and rotational kinetic energy.

Thus,

mgh = (1/2) mv² + (1/2) Iω²

Moment of Inertia for hoop; I = mr² Also, angular velocity; ω = v/r

Thus,

mgh = (1/2) mv² + (1/2) (mr²) (v/r) ² = mgh = (1/2) mv² + (1/2) mv²

mgh = mv²

The m will cancel out and so;

gh = v²

9.81 (1.80) = v²

v² = 17.658

v = √17.658

v = 4.2 m/s

4.24m/s

Explanation:

Potential energy at the top = kinetic energy at the button

But kinetic energy = sum of linear and rotational kinetic energy of the hoop

PE = mgh

KE = 1/2 mv^2

RE = 1/2 I ω^2

Where

m = mass of the hoop

v = linear velocity

g = acceleration due to gravity

h = height

I = moment of inertia

ω = angular velocity of the hoop.

But

I = m r^2 for hoop and ω = v/r

giving

m g h = 1/2 m v^2 + 1/2 (m r^2) (v^2/r^2) = 1/2 m v^2 + 1/2 m v^2 = m v^2

and m's cancel

g h = v^2

Hence

v = √gh

v = √10*1.8

v = 4.24m/s