Distance rolled of sphere vs cylinder with same m, r, and v

[I_Try_Math]

Homework Statement

A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height of 1.0 m. If a hollow sphere of the same mass and radius is given the same initial velocity, how high vertically does it roll up the incline?

Relevant Equations

$K_T = \frac{1}{2}mv^2$
$K_R = \frac{1}{2}Iw^2$

The answer from the textbook is:
Use energy conservation
$\frac{1}{2}mv^2 + \frac{1}{2}I_{cyl}w^2 = mgh_{cyl}$
$\frac{1}{2}mv^2 + \frac{1}{2}I_{sph}w^2 = mgh_{sph}$

Subtracting the two equations, eliminating the initial translational energy, we have:

$h_{cyl} = \frac{v^2}{g}$
$h_{sph} = \frac{5v^2}{6g}$

$h_{cyl} - h_{sph} = 0.43$ m

Thus, the hollow sphere, with the smaller moment of inertia, rolls up to a lower height of $1 - 0.43 = 0.57$ m.

I'm having trouble understanding the answer. If $h_{cyl} = \frac{v^2}{g}$, then doesn't the cylinder reach a vertical height of 2.55 m? Which would be a contradiction of the original premise of the question (that the cylinder rolls to a vertical height of 1 m)?

 

[haruspex]

Subtracting the two equations, eliminating the initial translational energy, we have:

$h_{cyl} = \frac{v^2}{g}$
$h_{sph} = \frac{5v^2}{6g}$

I don't see how that would be a useful step to take, nor how it could produce those equations. You need to cancel all the initial energy.

 

[andrewkirk]

You are correct.
The question does not need to specify the initial velocity. It only needs to say that the sphere is given the same velocity. Given the height to which it rolls, we can infer the initial velocity, and we would infer a smaller velocity than 5 m/s.
The question should not have stated the initial velocity. The author probably didn't notice they'd used an incorrect velocity because the number is not used anywhere in the calc, so a wrong number can't make it go wrong.
However your calc of 0.43m is wrong. Look at the two equations above that, substitute 1m for $h_{cyl}$ and redo the calc. The difference in heights is much less than 0.43.
And as haruspex has just pointed out your two equations cannot be derived by the step you describe (although they can be derived from other, different steps).

 

[I_Try_Math]

You are correct.
The question does not need to specify the initial velocity. It only needs to say that the sphere is given the same velocity. Given the height to which it rolls, we can infer the initial velocity, and we would infer a smaller velocity than 5 m/s.
The question should not have stated the initial velocity. The author probably didn't notice they'd used an incorrect velocity because the number is not used anywhere in the calc, so a wrong number can't make it go wrong.
However your calc of 0.43m is wrong. Look at the two equations above that, substitute 1m for $h_{cyl}$ and redo the calc. The difference in heights is much less than 0.43.
And as haruspex has just pointed out your two equations cannot be derived by the step you describe (although they can be derived from other, different steps).

Thank you for that explanation. So if I understand correctly, the answer should be that the sphere rolls 0.83 m.

 

[haruspex]

Thank you for that explanation. So if I understand correctly, the answer should be that the sphere rolls 0.83 m.

Yes.
Wrt the 5.0m/s, maybe this is not on Earth.

 

[Chestermiller]

What is the equation for the moment of inertial of a hollow sphere in terms of its mass, thickness, and outer radius?