Calculating Average Acceleration on a Ferris Wheel

[ffrpg]

The problems reads, A carnival Ferris wheel has a 14.2m radius and completes one revolution about its horizontal axis in 10.1s. Consider the time interval from the instant when a passenger is at the lowest point of the uniform circular motion to a clock reading 5.05 seconds later. What was the magnitude of the passenger's average acceleration during this time interval?


Do I use the formula (they're suppose to be vectors) a=(-rwcoswt)i+(-rwsinwt)j? I really don't have any direction towards solving this problem. Any hints would be wonderful.

 

[quantumdude]

Originally posted by ffrpg
Do I use the formula (they're suppose to be vectors) a=(-rwcoswt)i+(-rwsinwt)j? I really don't have any direction towards solving this problem. Any hints would be wonderful.

That should be:

a=-rω²cos(ωt)i-rω²sin(ωt)j

You forgot to square the angular velocity, but other than that it's OK.

 

[M98Ranger]

To calculate the average acceleration of the passenger on the Ferris wheel, we can use the formula a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity, and t is the time interval. In this case, the passenger's initial velocity is zero, since they start at the lowest point of the Ferris wheel, and their final velocity is also zero, since they return to the same point after one revolution.

So, we can rewrite the formula as a = (0 - 0)/5.05 = 0 m/s^2. This means that the passenger's average acceleration during this time interval is zero, indicating that their speed and direction did not change.

To answer your question about using the formula a = (-rwcoswt)i + (-rwsinwt)j, this formula is used to calculate the instantaneous acceleration at a specific point on the Ferris wheel, taking into account the position, velocity, and angular velocity of the wheel. In this problem, we are looking for the average acceleration over a time interval, so we do not need to use this formula.

I hope this helps! Let me know if you have any further questions.