Comparing the kinetic energies among a solid sphere, a cylinder and a hoop

[hidemi]

Homework Statement

A solid sphere, a solid cylinder, and a hoop all have the same mass and radius. Each are sent down identical inclined planes starting from rest. Their kinetic energies at the bottom of the incline are Ksphere, Kcylinder, and Khoop. Which of the following is true?

a. Ksphere > Kcylinder

b. Khoop > Ksphere

c. Khoop > Kcylinder

d. Kcylinder > Khoop

e.No answer above is correct.

The answer is e,

Relevant Equations

KE = 1/2 * I* ω^2

I did the question as attached, so I think A and D are correct but the given answer is E.
Where am I wrong? Thanks.

 

[PeroK]

What do you think kinetic energy means in this question?

 

[Steve4Physics]

Where am I wrong? Thanks.

In addition to the other replies:
- you have ignored the translational kinetic energy ($\frac 1 2 mv^2$);
- you have overcomplicated a simple problem - no formulae/maths are needed.

 

[haruspex]

- you have ignored the translational kinetic energy ($\frac 1 2 mv^2$);

In the working, the motion is treated as rotation about point of contact, so includes linear.

 

[TSny]

Did you mean for this to be $\omega \, ' ^2 = \omega^2 + 2\alpha \theta$?

Note in particular the squares on the angular speeds.

[Edit: the above is to help you identify your mistakes. As @Steve4Physics noted, no calculation is actually necessary to answer the question.]

 

[haruspex]

I might be missing something, but I would think about conservation of energy.

Certainly, but it is also useful for the OP to identify the error in the approach used.

 

[Steve4Physics]

In the working, the motion is treated as rotation about point of contact, so includes linear.

You're quite right. I should have read the given solution properly!

 

[hidemi]

Thanks for all your comments! I need some time to think about it.

 

[JLT]

The same starting potential energy turns into the same kinetic energy at the bottom: Ksphere = Ksolid cylinder = Khoop; K = K = K, they all have equal energies. same mgh → same kinetic energy

they do have different translational and rotational velocities due to their different I's:
https://en.wikipedia.org/wiki/List_of_moments_of_inertia
Larger I = more difficult to rotate.
Larger I's = more of that Kinetic energy is sucked into rotational motion rather than translational motion. The hoop rotates more slowly, but the total energies are the same.

 

[haruspex]

they do have different translational and rotational velocities due to their different I's:
https://en.wikipedia.org/wiki/List_of_moments_of_inertia
Larger I = more difficult to rotate.
Larger I's = more of that Kinetic energy is sucked into rotational motion rather than translational motion. The hoop rotates more slowly, but the total energies are the same.

You could save yourself some typing if you were to read the attempt first.

 

[JLT]

Homework Statement:: A solid sphere, a solid cylinder, and a hoop all have the same mass and radius. Each are sent down identical inclined planes starting from rest. Their kinetic energies at the bottom of the incline are Ksphere, Kcylinder, and Khoop. Which of the following is true?

a. Ksphere > Kcylinder

b. Khoop > Ksphere

c. Khoop > Kcylinder

d. Kcylinder > Khoop

e.No answer above is correct.

The answer is e,
Relevant Equations:: KE = 1/2 * I* ω^2

I did the question as attached, so I think A and D are correct but the given answer is E.
Where am I wrong? Thanks.

A couple things just a little off in your written work, need w^2 for constant acceleration equations, and moment balance needs to add to both rXma and I*alpha:
Here are two different ways to walk around it, using an energy balance, and also using constant acceleration equ's with moment balance, both reduce to the same results.

↑↑ different rotational velocities and angular accelerations, but it all adds up to the same mgh.

If given the choice in how to solve a problem, energy balance is usually the easiest way to go as energy is a scalar property (you do not have to worry about directions, only magnitudes, when it comes to 0.5mv^2 etc.)
Great job thinking through the relative magnitudes of the rotational motions! I think this questions was really trying to get you to think in terms of energy balances though.

 

[haruspex]

A couple things just a little off in your written work, need w^2 for constant acceleration equations, and moment balance needs to add to both rXma and I*alpha:
Here are two different ways to walk around it, using an energy balance, and also using constant acceleration equ's with moment balance, both reduce to the same results.
View attachment 279429
↑↑ different rotational velocities and angular accelerations, but it all adds up to the same mgh.

If given the choice in how to solve a problem, energy balance is usually the easiest way to go as energy is a scalar property (you do not have to worry about directions, only magnitudes, when it comes to 0.5mv^2 etc.)
Great job thinking through the relative magnitudes of the rotational motions! I think this questions was really trying to get you to think in terms of energy balances though.

AS @TSny spotted, the error was in writing $\omega=2\alpha S$, where presumably S was meant to be angular distance, so would be better written as θ.
Using $\omega^2=2\alpha \theta$ would have given the right result.

I previously interpreted the error differently. I thought the S was for time (seconds), and that the mistake was assuming time was the same for all three. But then the factor of 2 would have been wrong as well.