Finding the tangent acceleration of a car going around a loop.

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SUMMARY

The discussion focuses on calculating the tangential acceleration of a car at point C while navigating a loop. Given the height of h = 75 m and a radius R = 25 m, the velocity at point C is determined to be 31.3 m/s, with a radial acceleration of 39.2 m/s². The net acceleration calculated is approximately 40.45 m/s², derived from the forces acting on the car, specifically using principles from energy conservation, including mgh and 1/2mv². The participant acknowledges the role of gravitational acceleration in the calculations.

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btbam91
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Alright, I'm having trouble with this problem, but I know that it's easy and simple.

If the car starts at height h= 3.00 R and the radius is R_1 = 25.0 m, compute the tangential acceleration of the passengers when the car is at point C, which is at the end of a horizontal diameter.
Picture:

http://session.masteringphysics.com/problemAsset/1041727/8/YF-07-32.jpg


So h= 75 m and R = 25 m.

I know that the tangent acceleration is the derivative of the velocity equation with respect to time, but I can't find out how to do that exactly.

In early parts of this problem, I found the velocity at point C to be 31.3 m/s and the radial acceleration to be 39.2 m/s^2

This problem was found in my energy homework, so I know that I'm messing with mgh and 1/2mv^2 and such.

Thanks!
 
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There are two types of forces acting and by the second law, the net acceleration is just due to those two forces. Only one force is providing the tangential acceleration.

(I also found net acceleration to be 40.4476661 m/s^2 .. plugging numbers to the google calculator)
 
Acceleration due to gravity... boy do I feel silly right now.

Thanks bud!
 

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