Calculating the Minimum Height for a Safe Amusement Park Ride: A Physics Problem

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Homework Help Overview

The problem involves calculating the minimum height for a car on an amusement park ride to safely complete a loop without falling off. It is situated within the context of energy conservation and centripetal acceleration in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between potential energy (PE) and kinetic energy (KE) at different points of the ride, particularly focusing on the conditions at points A and B. Questions arise regarding the correct value of height h in relation to the radius R of the loop, and whether the energy conservation equation is applicable in this scenario.

Discussion Status

Some participants have provided insights into the energy transformations involved, suggesting that the mechanical energy at point A should equal the mechanical energy at point B. There is an ongoing exploration of the necessary conditions for the car to maintain contact with the track at the top of the loop, with no explicit consensus reached on the minimum height h.

Contextual Notes

Participants are considering the effects of potential energy conversion to kinetic energy and the implications of centripetal acceleration at the top of the loop. There is mention of assumptions regarding energy losses and the need for sufficient height to ensure safe passage through the loop.

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A car in an amusement park ride rolls without friction around the track. It starts from rest at point A at a height h above the bottom of the loop. (a) What is the minimum value of h (in terms of R) such that the car moves around the loop without falling off at the top (point B)? (b) If [tex]h = 3.5R[/tex] and [tex]R = 30 m[/tex] compute the speed, radial acceleration, and tangential acceleration of the passengers when the car is at point C, which is at the end of the horizontal diameter. Show acceleration components in a diagram.

All I know is that at point A, the kinetic energy will be equal to 0. Do I have to use [tex]K_{1} + U_{1} + W_{other} = K_{2} + U_{2}[/tex]? Also does [tex]h = 2R[/tex]

Any help would be appreciated

Thanks
 
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yes, and at point B you will have kenetic and potential energy.

does h = 2r? no
 
Yes the mechanical energy at A will be equal to the mechanical energy at B of the car. At B the car will need some force to supply the centripetal acceleration. Its weight is a candidate - if it moves faster the track will need to supply an additional normal force to increase the needed centripetal force. When this normal force is zero at the top the car is just leaving the track. Therefore the minimum centripetal acceleration at the top will be g. Use the connection between speed and centipetal acceleration to determine what its minimum speed will be a the top.
 
In this case, the car must have enough PE to convert into KE at the bottom of the loop and back into PE as it moves around the loop.. Therefore:
[tex]change PE_i= change KE_f= change PE_f[/tex]
simplifying it, [tex]change PE_i= change PE_f[/tex], provided there is no energy loses, if there is, then add it in.. since you want the h to be as small as possible, we can assume that the car actually loses all its KE at the top of the loop. So all you have to do is to sub in the values and get the h at start.
 
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