What is its rotational kinetic energy?

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SUMMARY

The discussion focuses on calculating the rotational kinetic energy (KE) of a system consisting of a 240g ball and a 570g ball connected by a 48.0-cm massless rod, rotating at 110 rpm. The correct formula for rotational kinetic energy is KE = 0.5 * I * ω², where I is the moment of inertia. Participants emphasized the need to first determine the center of mass (COM) of the system and then calculate the moment of inertia for each mass about that COM using the parallel axis theorem. The final calculation of KE follows from these steps.

PREREQUISITES
  • Understanding of rotational dynamics and kinetic energy
  • Familiarity with the concept of center of mass (COM)
  • Knowledge of moment of inertia calculations
  • Ability to apply the parallel axis theorem
NEXT STEPS
  • Learn how to calculate the center of mass for a system of particles
  • Study the parallel axis theorem in detail
  • Explore the derivation of moment of inertia for various shapes, including solid spheres
  • Practice problems involving rotational kinetic energy calculations
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Students studying physics, particularly those focusing on mechanics and rotational motion, as well as educators looking for examples of rotational kinetic energy problems.

Elleboys
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Homework Statement


A 240g ball and a 570g ball are connected by a 48.0-cm-long massless, rigid rod. The structure rotates about its center of mass at 110 rpm.



Homework Equations


KE = Iω^2
I = 1/12(mr^2)


The Attempt at a Solution


Since it has two masses and two different radius, I was not sure with what I should've done.
 
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Elleboys said:

Homework Statement


A 240g ball and a 570g ball are connected by a 48.0-cm-long massless, rigid rod. The structure rotates about its center of mass at 110 rpm.



Homework Equations


KE = Iω^2
I = 1/12(mr^2)


The Attempt at a Solution


Since it has two masses and two different radius, I was not sure with what I should've done.

Where did your formula for the moment of inertia come from? That doesn't look like the moment of inertia of a solid sphere (ball) to me.

To do this problem, you need the radius of each ball. Are you given this?

The first step is to find the common centre of mass (COM) of the system. Do you know how to do that?

The next step is to calculate the moment of inertia of the system about that common COM. This can be done by summing up the moments of each ball. Remember to use the right formula and remember the parallel axis theorem.

The final step to find the rotational KE is trivial. But you should note that even your formula for rotational KE is wrong (missing a factor of 0.5).
 
Last edited:
Curious3141 said:
Where did your formula for the moment of inertia come from? That doesn't look like the moment of inertia of a solid sphere (ball) to me.

To do this problem, you need the radius of each ball. Are you given this?

The first step is to find the common centre of mass (COM) of the system. Do you know how to do that?

The next step is to calculate the moment of inertia of the system about that common COM. This can be done by summing up the moments of each ball. Remember to use the right formula and remember the parallel axis theorem.

The final step to find the rotational KE is trivial. But you should note that even your formula for rotational KE is wrong (missing a factor of 0.5).

OHHHHH I see
And yes I put wrong formula for I.
So I believe that I need to find a center of mass, get moment of inertia of each particle about that COM, add them up and it will give me net moment of inertia.
And by using KE = Iω^2, I can get the answer.
Am I on the right track?
 
Elleboys said:
And by using KE = Iω^2, I can get the answer.
Am I on the right track?

You still need to add the 0.5 on front of your formula for KE like Curious3141 said. KE = 0.5Iω^2
 
Elleboys said:
OHHHHH I see
And yes I put wrong formula for I.
So I believe that I need to find a center of mass, get moment of inertia of each particle about that COM, add them up and it will give me net moment of inertia.
And by using KE = Iω^2, I can get the answer.
Am I on the right track?

Yes, you're on the right track (except that ##K = \frac{1}{2}I\omega^2##). Work through it systematically.
 

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