Total acceleration at the bottom of a rolling wheel

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Discussion Overview

The discussion revolves around calculating the total acceleration at the bottom of a rolling wheel, specifically in the context of a tundra buggy stuck on slippery ice. Participants explore the components of acceleration, including tangential, radial, and linear accelerations, while addressing the implications of wheel slipping on the calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states the total acceleration is the sum of tangential, radial, and linear accelerations but struggles to find the correct values for these components.
  • Another participant questions how speedometers function and what they actually measure, suggesting it is related to angular speed rather than overland speed.
  • A participant proposes that angular acceleration can be calculated as final angular velocity divided by time, but expresses concern that the resulting value seems small.
  • It is noted that linear acceleration is radius times angular acceleration, but this relationship may not hold if the wheel is slipping on the ice.
  • One participant emphasizes that speedometers are calibrated to measure the tangential speed of the wheel rim, assuming no slipping occurs.
  • There is a suggestion to consider whether acceleration remains constant throughout the motion and the role of distance moved in the calculations.
  • Participants hint at the need for additional notes regarding rolling with slipping, indicating a potential complexity in the scenario.

Areas of Agreement / Disagreement

Participants express differing views on the calculations of acceleration components and the implications of slipping, indicating that the discussion remains unresolved with multiple competing perspectives.

Contextual Notes

There are limitations regarding the assumptions made about slipping and the constancy of acceleration, which are not fully addressed in the discussion.

Angelique
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A tundra buggy, which is a bus fitted with oversized wheels, is stuck in Churchill, Manitoba, on slippery ice. The wheel radius is 0.84 m. The speedometer goes from 0 to 27 km/h while the buggy moves a total distance of 7.0 m in 9.0 s.

Find the magnitude of the total acceleration of a point at the bottom of the wheel at the end of 4.0 s.

I know that the total acceleration is the tangential acceleration + the radial acceleration + the linear acceleration. But i can't seem to get the right values for the accelerations since the answer is 13.2m/s^2

for tangential acceleration i got 0.66m/s^2
Im not sure how to find the other 2 accelerations
 
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How do speedos work?
What does the speedo actually measure? (Hint: not the overland speed.)
 
It measures the speed of the wheel (angular speed) so then the angular acceleration is: (final angular velocity)/time? so 7.5/8 = 0.9375m/s^2? but that seems very small to me because linear acceleration is radius times the angular acceleration. :/
 
Angelique said:
linear acceleration is radius times the angular acceleration.
Not when the wheel is slipping on the ice.
 
It measures the speed of the wheel (angular speed) so then the angular acceleration is: (final angular velocity)/time? so 7.5/8 = 0.9375m/s^2? but that seems very small to me because linear acceleration is radius times the angular acceleration. :/
The speedo reads the vehicle speed in the event there is no slipping.
So it is calibrated to tell you the tangential speed of the wheel rim.
So the outer edge of the wheel goes from 0 to v in time T, giving an average angular acceleration of ##\alpha_{ave} = v/rT##
... notice that ##a_T=r\alpha=v/T## as expected?
(It is best practice to do the algebra with symbols and substitute the numbers in later.)

Note: that's just an example - you may need to consider more than the average acceleration. i.e. is the acceleration the same throughout the motion? What part does the distance moved play?
I'm guessing you have some notes about rolling with slipping?
 
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