Deriving work-kinetic energy theorem for rotational motion

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SUMMARY

The work-kinetic energy theorem for rotational motion is derived using the equation W=(0.5*I)*((ωf^2)-(ωi^2)), where W represents work, I is the moment of inertia, and ωf and ωi are the final and initial angular velocities, respectively. The derivation starts from W=τ*(θ), with τ defined as torque. The correct relationship between torque and force is τ = r*F, which must be used to avoid errors in calculations. The discussion highlights the importance of understanding the definitions and relationships between torque, force, and angular motion.

PREREQUISITES
  • Understanding of rotational dynamics, specifically torque and moment of inertia.
  • Familiarity with angular velocity and angular acceleration concepts.
  • Knowledge of basic kinematic equations for rotational motion.
  • Ability to apply calculus in physics problems, particularly in deriving relationships.
NEXT STEPS
  • Study the derivation of the work-energy theorem in rotational motion.
  • Learn about the relationship between torque and angular momentum.
  • Explore the application of the cross product in rotational dynamics.
  • Investigate the concept of moment of inertia for various shapes and its implications in rotational motion.
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators teaching rotational dynamics and the work-energy theorem.

gimak
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Homework Statement



How to derive W=(.5*I)*((ωf^2)-(ωi^2)) from W=τ*(θ)

Homework Equations



The two equations seen above (big W is work; little w is rotational velocity, τ is torque, alpha=rotational acceleration).

s=r*θ a=r*(alpha) (ωf^2)-(ωo^2)=2*(alpha)*(θ) F=r*τ F=ma

The Attempt at a Solution



W=T*(θ)=(F/r)*(s/r)=(r^-2)*F*s=(r^-2)*ma*s=(r^-2)*m*(r*alpha)((ωf^2)/(2*alpha))*r
=0.5*m*ωf^2

Obviously this isn't right. What went wrong?
 
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Are you allowed to use calculus?
 
gimak said:

Homework Statement



How to derive W=(.5*I)*((ωf^2)-(ωi^2)) from W=τ*(θ)

Homework Equations



The two equations seen above (big W is work; little w is rotational velocity, τ is torque, alpha=rotational acceleration).

s=r*θ a=r*(alpha) (ωf^2)-(ωo^2)=2*(alpha)*(θ) F=r*τ
[/QUOTE]

The formula in red is wrong. How is the torque defined?

ehild
 
response to ehlid

How is it wrong? Force is torque times radius, is it not. Oh, I can't use scalar multiplication on this. Do I have to use the cross product: F=t x r?
 
Or τ = I*α = I*a/r maybe...

and you should think of your s as Δs
 
gimak said:
How is it wrong? Force is torque times radius, is it not.

It is not. Check your textbooks.

ehild
 
correction

Shoot! I'm sorry! T=r*F. I got it backwards! If I plug in the right equation, the first equation will eventually turn into 0.5*m*(r^2)*(ωf^2)=I*(ωf^2).
 
Last edited:

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