How Much of a Bicycle's Kinetic Energy is in Its Wheels?

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SUMMARY

The discussion focuses on calculating the proportion of a bicycle's kinetic energy that resides in its wheels, specifically addressing a bicycle weighing 23lbf with hollow wheels of 2.5lbf each. The correct approach involves using the equations for translational kinetic energy (KE = 1/2mv^2) and rotational kinetic energy (Ke = I(omega)^2), where I = mr^2 for each wheel. The conclusion reached is that 36% of the total kinetic energy is attributed to the wheels, emphasizing the importance of considering both translational and rotational components in the calculation.

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  • Understanding of basic physics concepts, including kinetic energy and inertia.
  • Familiarity with the equations for translational and rotational motion.
  • Knowledge of how to calculate moment of inertia for rotating objects.
  • Ability to perform calculations involving weight and energy in imperial units.
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  • Review the principles of rotational dynamics, focusing on moment of inertia calculations.
  • Study the relationship between translational and rotational kinetic energy in rigid bodies.
  • Learn how to apply the parallel axis theorem for complex systems.
  • Explore practical applications of kinetic energy calculations in bicycle design and performance optimization.
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This discussion is beneficial for physics students, mechanical engineers, and bicycle enthusiasts interested in understanding the dynamics of bicycle motion and energy distribution in moving systems.

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Homework Statement



A bicycle weighs 23lbf, including hollow wheels each weighing 2.5lbf. Calculate how much of the bike's kinetic energy is in the wheels. Ans = 36%

Homework Equations


1. I = mr^2 <----- i think this is 2mr^2 because of two wheels
2. Kinetic Energy = 1/2mv^2 <---- translational of the bike
3. Ke = I(omega)^2 <------ rotational of the wheels
4. Ktotal = 1/2mv^2 + Iw^2

The Attempt at a Solution



I tried putting the rotational kinetic energy over the total kinetic then multiplying by 100 but it isn't working... Also I am not sure if I am supposed to double the rotational energy or not. I think that I am only supposed to double the inertia. Either way i can't seem to arrive at the given solution. Am i doing something wrong?
 
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The wheels carry energy in two ways as you suggest. I would think then it would be 2*(1/2mv^2+Mr^2)/ (the translational kinetic energy plus the rotational of the whole bike)--perhaps you neglected the latter term in your calculation?
 

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