Deriving work-kinetic energy theorem for rotational motion

In summary, the conversation is about deriving the formula W=(.5*I)*((ωf^2)-(ωi^2)) from W=τ*(θ), with the equations for work, torque, rotational velocity, rotational acceleration, displacement, and force. The attempt at the solution involved using the formula W=T*(θ), but it was incorrect because force is not equal to torque times radius. The correct equation should be T=r*F. The conversation also mentioned using calculus and considering Δs instead of s.
  • #1
gimak
56
1

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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  • #2
Are you allowed to use calculus?
 
  • #3
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
 
  • #4
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?
 
  • #5
Or τ = I*α = I*a/r maybe...

and you should think of your s as Δs
 
  • #6
gimak said:
How is it wrong? Force is torque times radius, is it not.

It is not. Check your textbooks.

ehild
 
  • #7
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:

1. What is the work-kinetic energy theorem for rotational motion?

The work-kinetic energy theorem for rotational motion states that the work done on an object rotating about a fixed axis is equal to the change in its kinetic energy. This means that the net torque applied to the object is equal to the change in its moment of inertia multiplied by its angular acceleration.

2. How is the work-kinetic energy theorem derived for rotational motion?

The work-kinetic energy theorem for rotational motion can be derived by integrating the equation for torque, τ = Iα, with respect to time. This yields the equation W = ΔKE, where W represents the work done on the object and ΔKE represents the change in its kinetic energy.

3. What are the key assumptions made in deriving the work-kinetic energy theorem for rotational motion?

The key assumptions made in deriving the work-kinetic energy theorem for rotational motion are that the object is rigid, the axis of rotation is fixed, and there is no external force acting on the object other than the torque causing its rotation.

4. Can the work-kinetic energy theorem be applied to all types of rotational motion?

Yes, the work-kinetic energy theorem can be applied to all types of rotational motion, including both translational and rotational motion. This is because the theorem relates the work done to the change in kinetic energy, which is a scalar quantity and therefore independent of the direction of motion.

5. How is the work-kinetic energy theorem used in real-world applications?

The work-kinetic energy theorem is used in various real-world applications, such as in the design of mechanical systems, such as engines and motors, where rotational motion is involved. It is also used in the analysis of rotational motion in sports, such as in the throwing of a discus or the swinging of a golf club.

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