Frame of reference of bicycle rider

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SUMMARY

The discussion focuses on the physics of a bicycle wheel, specifically calculating angular speed, momentum, angular momentum, and kinetic energy from the rider's frame of reference. The wheel has a mass of 1 kg and a diameter of 0.7 m, leading to an angular speed (ω) of 10 rad/s. The momentum (p) is 0 kg·m/s due to the lack of linear translation in the rider's frame. The angular momentum (L) is calculated as 1.225 kg·m², and the kinetic energy of the wheel is determined to be 6.125 J.

PREREQUISITES
  • Understanding of angular motion and linear motion concepts
  • Familiarity with the equations of motion for rotating bodies
  • Knowledge of moment of inertia, particularly for a hoop
  • Basic principles of kinetic energy in rotational dynamics
NEXT STEPS
  • Study the derivation of the moment of inertia for different shapes
  • Learn about the relationship between linear and angular momentum
  • Explore the effects of friction and air resistance on bicycle dynamics
  • Investigate energy conservation in rotational systems
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in the dynamics of rotating objects, particularly in the context of bicycles.

Alex_Neof
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Homework Statement


A wheel of a bicycle weighs 1 kg, all of which is at the rim. The diameter of the wheel is 0.7 m. If the bicycle is ridden at the speed of 3.5 m s–1, then in the frame of
reference of the rider:

i) Calculate the angular speed ω of the wheel.

ii) Calculate the momentum of the rim, p.

iii) Calculate the angular momentum of the wheel.

iv) Calculate the kinetic energy of the wheel.

Homework Equations



## v = {\omega} r ##
## \rho = mv ##
## L=I\omega ##
## Kinetic\ Energy = \frac{1}{2} m v^2 + \frac{1}{2} I \omega ^2##

Moment of Inertia of a hoop (for the wheel, ignoring the spokes):

## I=M R^2 ##

The Attempt at a Solution


[/B]
i) ## \frac {v}{r} = 10\ rad\ s^{-1}##

ii) ## 0 \ {kg}\ m\ s^{-1}## (Since in the frame of reference of the rider, there is no linear translation?)

iii) ## L=(M R^2) \omega ##
##(1) (0.35)^2 (10) = 1.225 {kg}\ m^2 ##

iv) ##Kinetic\ Energy = \frac{1}{2} I \omega ^2## Ignoring the linear term again.

##Kinetic\ Energy = (0.5) (0.35) ^2 (10)^2 = 6.125 J ##

Is this correct?

Thank you.
 
Last edited:
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Looks right.
 
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