Don't hula hoop on a hill

The potential energies of both objects is the same initially. When they are at the bottom of the hill it is easy to show their kinetic energy is then $\frac{1+I/mR^2}{2}v^2$ and is equal to $5mg$ . But then substituting the moment of inertias we get the ratio between the squares of velocities is $4/3$ , and thus the ratio is about $1.15$ which is the answer.

[Extended note from David: This solution has the physics correct and no unnecessary steps, which is great. However, keep in mind that being correct is only the first step to success in a technical field. Also key is whether you can explain what's going on to others. Good expository skills are not just for those interested in teaching, they are necessary no matter what profession you go into. Engineering firms want people who can work in teams and communicate their ideas, doctors need to explain medical issues to their patients, and even at the highest levels of academia scientists still communicate via papers and talks. A great idea that you can't explain doesn't do anyone any good.

So, what could be improved in this solution in terms of the exposition? Let's look at the second sentence:

``When they are at the bottom of the hill it is easy to show their kinetic energy is then $\frac{1+I/mR^2}{2}v^2$ and is equal to $5mg$ .''

Here we have a little mixed notation. In some places the m is still in the equation, in others it has been factored out and vanished since it is numerically equal to 1. When writing solutions, try to write out equations algebraically first and define the symbols in each equation before starting to substitute in numbers. As it's written above, it doesn't even look as if the equation has the right dimensions! Additionally, I, the moment of inertia, is never actually defined. Whereas someone who solved the problem wouldn't need a definition of I, the point of a solution is to explain to people who don't understand the problem how to do it. Finally, when the statement that the kinetic energy is equal to $5mg$ is made, why is that? It's true, but the reason is conservation of energy. In physics it's not the equations that are important, but the physical principles behind them. Those should be stated in solutions in general.]

In both cases energy is conserved. Hence we have that the total translational kinetic, rotational kinetic, and potential energy is a constant,

$\frac{1} {2} m v_{cm}^2 + \frac{1} {2} I \omega^2+U_g=constant=E_i$ ,

where $v_{cm}$ is the velocity of the center of mass, I is the moment of inertia, $\omega$ is the angular velocity, $U_g$ is the gravitational potential, and $E_i$ is the initial total energy.

The initial potential energy is $mg(5.5)$ , the final potential energy is $mg(0.5)$ , and the initial translational and rotational kinetic energies are zero. Therefore $E_i=53.9$ J. Since the hoop and disk roll without slipping, we can relate $\omega=v_{cm}/R$ , where R is the radius of the hoop/disk. The moment of inertia of a hoop is $I=mR^2$ while that of a disk is $I=\frac{1} {2} mR^2$ . Putting it all together allows us to solve for $v_{cm}$ at the bottom of the hill for both the hoop and the disk, and the ratio $v_d/v_h$ evaluates to 1.15.

Since both CoG fall through same height, the change in PE is the same. From conservation of energy, change in PE = KE acquired. So, at the end, both will have the same KE. So we have

$KE =~\dfrac{1}{2}I*v^2+\dfrac{1}{2}I*\omega^2 \\ I= k* \dfrac{1}{2}m*R^2.~~\\Since ~they ~roll~ without~ slipping~\omega*R = v. ~So~ k*I=k*m*v^2.\\\therefore~KE=\dfrac{1}{2}m*v^2+\dfrac{1}{2}k*m*v^2\\ \implies~KE=\dfrac{1}{2}m*v^2*(1+k)\\For~ Hoop~I_h=m*R^2, ~~~~\therefore~ k_h=1\\For~disc~I_d=\dfrac{1}{2}*m*R^2,~~~~~ \therefore~ k_h=\dfrac{1}{2}\\ But ~KE_h=KE_d ~\\ \implies~ KE_h=\dfrac{1}{2}m*v_h^2*(1+1)= KE_d=\dfrac{1}{2}m*v_d^2*(1+\dfrac{1}{2})\\ \implies~2*v_h^2=\dfrac{3}{2}*v_d^2\\ \therefore\dfrac{v_d}{v_h}=\dfrac{2}{\sqrt3}=\boxed{1.1547}$

This can be solved even if the numeric values of masses and radius of ring were not given.

If we draw the free body diagram of disc and ring(hula hoop) we will come to know that there are three forces acting on the object. They are

1) mg (vertically downward) 2) Normal force from the hill 3)Frictional force.

Using newton's second law we will get $mgsin\theta -f=ma$ ..................(1)

Rotational equation

$fr=I\alpha$ ................(2)

As the object are performing pure rolling motion

So $a=r\alpha$ .................(3)

from eq(2) and (3)

$f=\frac { Ia }{ { r }^{ 2 } }$

Putting this value in eq(1) we will get

$mgsin\theta -\frac { Ia }{ { r }^{ 2 } } =ma$

For ring $I=m{ r }^{ 2 }$ and for disc $I=\frac { m{ r }^{ 2 } }{ 2 }$

Putting these values in eq(1) and $\theta=30$ we will get acceleration of ring = g /4 and acceleration of disc= g/3

Distance covered by object by object along the hill is $5.5/sin\theta$ which is equal to 11

As ${v}^{2} = {u}^{2} + 2as$

Putting the values of acceleration of ring and disc one by one we will find

${ v }_{ d }=\sqrt { \frac { 2g\times 11 }{ 3 } }$ and

${ v }_{ h }=\sqrt { \frac { g\times 11 }{ 2 } }$

So $\frac { { v }_{ d } }{ { v }_{ h } } =\frac { 2 }{ \sqrt { 3 } }$

:D.