Frame of reference of bicycle rider

AI Thread Summary
The discussion focuses on calculating various physical properties of a bicycle wheel from the rider's frame of reference. The angular speed of the wheel is determined to be 10 rad/s, while the momentum of the rim is calculated as 0 kg m/s due to the lack of linear translation in the rider's frame. The angular momentum of the wheel is found to be 1.225 kg m², and the kinetic energy is calculated to be 6.125 J, excluding linear kinetic energy. The calculations are confirmed as correct by participants in the discussion. The thread effectively illustrates the application of physics concepts to a practical scenario involving a bicycle wheel.
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.
 
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Looks right.
 
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