Total kinetic energy for rolling hoop

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SUMMARY

The total kinetic energy of a rolling hoop with a mass of 200g and a radius of 25cm, moving at a velocity of 5m/s, is calculated to be 5J. The moment of inertia (ICM) is determined using the formula ICM = MR², resulting in a value of 0.0125 kg·m². The rotational kinetic energy is calculated using K = 1/2Iω², where ω is derived from the relationship v = rω, yielding a rotational energy of 2.5J. The total kinetic energy must account for both translational and rotational motion, which is crucial for accurate calculations.

PREREQUISITES
  • Understanding of rotational dynamics and kinetic energy equations
  • Familiarity with the moment of inertia formula (ICM = MR²)
  • Knowledge of the relationship between linear velocity and angular velocity (v = rω)
  • Basic principles of rolling motion without slipping
NEXT STEPS
  • Study the derivation of the total kinetic energy formula for rolling objects
  • Learn about the differences between translational and rotational kinetic energy
  • Explore examples of moment of inertia for various shapes
  • Investigate the effects of friction on rolling motion
USEFUL FOR

Physics students, educators, and anyone interested in understanding the principles of rotational dynamics and kinetic energy in rolling objects.

vetgirl1990
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Homework Statement


A hoop has a mass of 200g and a radius of 25cm (ICM = MR2 ). It rolls without sliding along level ground at VCM = 5m/s. What is it's total kinetic energy?

Homework Equations


K = 1/2Iω2
ICM = MR2
v = rω

The Attempt at a Solution


The answer is 5J, but I'm getting 2.5J.

ICM = MR2 = (0.2)(0.25)(0.25) 0.0125
ω = v / r = 5/0.25 = 20

K = 1/2(0.0125)(20)2 = 2.5J

Am I omitting something from the total kinetic energy equation?
 
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You should consider translational motion too.
 
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Likes   Reactions: vetgirl1990
azizlwl said:
You should consider translational motion too.

Thanks! I forgot that a rolling object experiences both translational AND rotational motion. D'oh!
 

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