Amusement Park Ride, Find Velocity at Point C

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SUMMARY

The discussion focuses on calculating the velocity of a car on an amusement park ride at point C, using the principle of conservation of energy. The car starts from rest at a height of h = 5.00 R, where R = 16.0 m. The equation derived is V = sqrt(2g(4R)), confirming that the gravitational acceleration g is used in the calculations. The participant successfully solved the problem and received confirmation on their approach, particularly the simplification of terms.

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Homework Statement



A car in an amusement park ride rolls without friction around the track shown in the figure (Figure 1) . It starts from rest at point A at a height h above the bottom of the loop. Treat the car as a particle.

If the car starts at height h= 5.00 R and the radius is R = 16.0m , compute the speed of the passengers when the car is at point C, which is at the end of a horizontal diameter.


Homework Equations



Conservation of Energy


The Attempt at a Solution



Using conservation of Energy

PE_i + KE_i = PE_f + KE_f

mgh = mgh + 1/2mv^2

mg(5R) = mg(R) + 1/2mv^2 (We are solving for V, so..)

mg(5R) - mg(R) = 1/2mv^2

(Cancel out mass)

g(5R) - g(R) = 1/2v^2

2(g(5R) - g(R)) = v^2

sqrt(2(g(5R) - g(R))) = V

(Plug in R)

sqrt(2(g(80) - g(5))) = V

If this is correct so far, is g just gravity or is it centripetal acceleration? I loose confidence at this step..Lead me in the right direction?

Thanks!
 

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Looks like you're doing alright.

Note that during your simplifications you could make mg(5R) - mg(R) into mg4R and save a lot of number pushing later on down the page.
 
gneill said:
Looks like you're doing alright.

Note that during your simplifications you could make mg(5R) - mg(R) into mg4R and save a lot of number pushing later on down the page.

Ok great, submitted my answer and it was right. Thanks for the tip for subtracting the R's. I didn't know if I was doing the right process. Thanks again
 

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