Question

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?

Answer 1

Answer:

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:

$PE=KE_{tra} + KE_{rot}$ (1)

where:

$PE=mgh$ is the initial potential energy of the hoop at the top of the hill, with

m = mass of the hoop

$g=9.8 m/s^2$ (acceleration of gravity)

$h=4.50 m$ (height of the hill)

$KE_{tra}=\frac{1}{2}mv^2$ is the translational kinetic energy, where

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

$KE_{rot}=\frac{1}{2}I\omega^2$ is the rotational kinetic energy, where

$I$ is the moment of inertia of the hoop

$\omega$ is the angular velocity

The moment of inertia of a hoop is given by

$I=mR^2$

where R is the radius of the hoop.

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

$\omega=\frac{v}{R}$

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

$KE_{rot}=\frac{1}{2}(mR^2)(\frac{v}{R})^2=\frac{1}{2}mv^2$

So, eq.(1) becomes:

$mgh=\frac{1}{2}mv^2+\frac{1}{2}mv^2$

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

$mgh=mv^2\\v=\sqrt{gh}=\sqrt{(9.8)(4.50)}=6.6 m/s$