Total acceleration at the bottom of a rolling wheel

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A tundra buggy in Churchill, Manitoba, is experiencing issues on slippery ice, with a wheel radius of 0.84 m and a speedometer reading that fluctuates from 0 to 27 km/h over a distance of 7.0 m in 9.0 seconds. The total acceleration at the bottom of the wheel at 4.0 seconds is calculated to be 13.2 m/s², but there is confusion regarding the values for tangential, radial, and linear acceleration. The speedometer measures the angular speed of the wheel rather than the actual vehicle speed, which complicates the calculations. It is noted that the linear acceleration is derived from the radius multiplied by the angular acceleration, but slipping affects this relationship. Understanding the dynamics of rolling with slipping is crucial for accurate calculations.
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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