Calculating Rotational Kinetic Energy of a Baton

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SUMMARY

The discussion focuses on calculating the rotational kinetic energy of a 94-cm-long, 500 g baton twirled at 150 rpm. The correct formula for rotational kinetic energy is KE_rot = 0.5 I w^2, where I is the moment of inertia and w is the angular velocity in rad/s. After correcting the moment of inertia calculation to I = (1/12)MR^2, the final rotational kinetic energy is confirmed to be 4.5 J. This calculation is essential for accurately determining the energy involved in the baton’s motion.

PREREQUISITES
  • Understanding of rotational dynamics
  • Familiarity with the moment of inertia formula I = (1/12)MR^2
  • Knowledge of angular velocity conversion from rpm to rad/s
  • Basic proficiency in physics equations related to kinetic energy
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  • Study the derivation and applications of the moment of inertia for various shapes
  • Learn about angular momentum and its relation to rotational kinetic energy
  • Explore the effects of different rotational speeds on kinetic energy
  • Investigate real-world applications of rotational dynamics in sports and engineering
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Physics students, educators, and anyone interested in understanding rotational motion and energy calculations in mechanical systems.

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Homework Statement


A drum major twirls a 94-cm-long, 500 g baton about its center of mass at 150 rpm. What is the baton's rotational kinetic energy?

Homework Equations



KE_rot = 0.5 I w^2

The Attempt at a Solution


w = (150rpm*2pi)/60 = 15.708rad/s

I = (1/12)*0.5kg * 0.94m = 0.039167kg*m^2

KE_rot = 0.5 I w^2

KE_rot = 0.5(0.039167)(15.708rad/s)^2
= 4.8J

says wrong answer.
Cant figure out what I am doing wrong
thanks for any help
 
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Take a look at the moment of inertia: you do kg times m and you obtain kg m^2.
 
thanks..ok forgot to square so I = (1/12)MR^2
I = (1/12) 0.5kg ((0.94m)^2) = 0.03682kg*m^2

so KE_rot = 0.5(0.03682)(15.708rad/s)^2 = 4.5J

(i just want to confirm correct before I submit answer again)
thanks for the help
 
I think it is correct.
 

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