Translational vs. Rotational Kinetic Energy

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SUMMARY

The discussion centers on the distinction between translational and rotational kinetic energy when calculating the work done to bring a resting cylinder to an angular speed of 8 rad/s. It is incorrect to use the translational kinetic energy formula (0.5mv²) with tangential velocity derived from angular speed because the tangential velocity varies across different points on the cylinder's surface. Instead, the rotational kinetic energy formula (KE = 0.5Iω²) is necessary, where I represents the moment of inertia, as it simplifies the calculation and accounts for the distribution of mass in the cylinder.

PREREQUISITES
  • Understanding of rotational dynamics
  • Familiarity with the moment of inertia (I)
  • Knowledge of angular velocity and its relationship to tangential velocity
  • Basic principles of kinetic energy
NEXT STEPS
  • Study the derivation of the moment of inertia for various shapes
  • Learn about the relationship between angular speed and tangential speed in rotating bodies
  • Explore the applications of rotational kinetic energy in real-world scenarios
  • Investigate the differences between translational and rotational motion in physics
USEFUL FOR

Students of physics, mechanical engineers, and anyone interested in understanding the principles of rotational motion and energy calculations.

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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