# Total acceleration at the bottom of a rolling wheel

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1. Nov 9, 2014

### Angelique

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

2. Nov 9, 2014

### Simon Bridge

How do speedos work?
What does the speedo actually measure? (Hint: not the overland speed.)

3. Nov 9, 2014

### Angelique

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. :/

4. Nov 9, 2014

### rcgldr

Not when the wheel is slipping on the ice.

5. Nov 9, 2014

### Simon Bridge

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?

Last edited: Nov 9, 2014