Translational vs. Rotational Kinetic Energy

AI Thread Summary
To calculate the work done in bringing a resting cylinder to an angular speed of 8 rad/s, it is incorrect to use the translational kinetic energy formula (0.5mv²) based on tangential velocity because the cylinder's tangential velocity varies with distance from the axis. Instead, the rotational kinetic energy formula, which incorporates the moment of inertia (I) and angular speed (ω), is more appropriate. This is because the rotational kinetic energy accounts for the distribution of mass throughout the cylinder, allowing for a more accurate calculation. While it is possible to sum the translational kinetic energy of each differential mass element, using the rotational approach simplifies the process. Ultimately, the rotational kinetic energy formula provides a straightforward and effective means to determine the work done in this scenario.
kash25
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Hi,
Suppose I am trying to find the work done in bringing a resting cylinder to an angular speed of 8 rad/s.
Why is it INCORRECT to find the corresponding tangential velocity at a point on the outer surface of the cylinder (using angular speed * radius = tangential speed) and use the translational (0.5mv^2) work-kinetic energy theorem?
Why MUST we use the rotational version with I and angular speed?
Thank you.
 
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Because there is rotational kinetic energy as well.
 
kash25 said:
Why is it INCORRECT to find the corresponding tangential velocity at a point on the outer surface of the cylinder (using angular speed * radius = tangential speed) and use the translational (0.5mv^2) work-kinetic energy theorem?
Realize that the tangential velocity depends on the distance from the axis--the cylinder does not have a uniform tangential velocity. But if you're willing to add up the translational KE of each piece (dm) of the cylinder, that's just fine. (You'll get the same answer.)

KE = Σ½dm v² = Σ½dm r²ω² = ½(Σdm r²)ω² = ½Iω²
Why MUST we use the rotational version with I and angular speed?
It's just much easier. :wink:
 
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