Force on coaster = ma(c) + ma(g) ?

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The discussion centers on calculating the force exerted by the track on a roller-coaster car at the bottom of a circular dip. A 500 kg car traveling at 16 m/s in a dip with a radius of 40 m results in a calculated force of 8.1 kN. The conversation also touches on how to determine the force at the top of the loop, noting that it involves subtracting the gravitational force from the centripetal force. Clarification is provided that the direction of the forces must be considered depending on the coaster's position relative to the track. The importance of drawing diagrams and applying Newton's second law is emphasized for accurate calculations.
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Homework Statement



A roller-coaster car has a mass of 500 kg when fully loaded with passengers. At the bottom of a circular dip of radius 40 m (as shown in the figure) the car has a speed of 16 m/s. What is the magnitude of the force of the track on the car at the bottom of the dip?

Homework Equations



Force on an object traveling in a circle perpendicular to the force of gravity is centripetal acceleration plus acceleration due to gravity times the mass of the object... i think

The Attempt at a Solution



16 meters per sec squared divided by 40 meters times 500 kg ... plus 500 kg times 9.8 meters per second squared = 8.1 kN ... right?
 
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Hi CaptainADHD! :smile:
CaptainADHD said:
Force on an object traveling in a circle perpendicular to the force of gravity is centripetal acceleration plus acceleration due to gravity times the mass of the object... i think
…
16 meters per sec squared divided by 40 meters times 500 kg ... plus 500 kg times 9.8 meters per second squared = 8.1 kN ... right?

Right! :biggrin:
 
tiny-tim said:
Hi CaptainADHD! :smile:


Right! :biggrin:

Great, thanks!

So if I wanted to calculate the force on the coaster at the top of the circle, it would be centripetal acceleration times mass MINUS the force of gravity, since gravity is in the opposite direction, right?
 
CaptainADHD said:
So if I wanted to calculate the force on the coaster at the top of the circle, it would be centripetal acceleration times mass MINUS the force of gravity, since gravity is in the opposite direction, right?

It depends whether the coaster is under the track or on top of it at the top of the loop …

draw two diagrams, and do the Newton's second law equation, and you'll see what I mean. :wink:
 
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