What is the final velocity (in m/s) of a hoop that rolls without slipping down a 4.50-m-high hill, starting from rest?

Given that:

By using the conservation of energy, we get

→

or,

→

By substituting the values, we get

Thus the response above is right.

Learn more about velocity here:

link

Final velocity: 6.6 m/s

Explanation:

As the hoop rolls down, its initial gravitational potential energy is converted into kinetic energy. However, since the hoop is rolling, its kinetic energy consists of both translational and rotational energy. So we can write:

(1)

where:

is the initial potential energy of the hoop at the top of the hill, with

m = mass of the hoop

(acceleration of gravity)

(height of the hill)

is the translational kinetic energy, where

v is the final speed of the hoop at the bottom of the hill

is the rotational kinetic energy, where

is the moment of inertia of the hoop

is the angular velocity

The moment of inertia of a hoop is given by

where R is the radius of the hoop.

Also, the angular velocity is related to the linear velocity by

Substituting the two last expressions into the expression for the rotational kinetic energy, we get

So, eq.(1) becomes:

And so we can solve it to find the final velocity of the hoop:

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )