At the top of the trajectory there are only 2 forces on Ș the centrifugal force directed outwoard the curve and the weight directed toward the centrer of the curve. The minimum speed is reached when these 2 forces are equal
m*v^2/R = m*g
hence Vminimum =sqrt(m*g/R)
At higer speeds the coaster is pushing on the track at lower speed the weight is higher than the centrifugal force. There is no normal force Fn as you have thinked of.
At what minimum speed must a roller coaster be traveling when upside down at the top of a circle if the passengers are not to fall out?
I'm having trouble understanding what it means by minimum speed. I know that at the top there are two forces, the force of gravity and the normal force, and both are in the direction of the centripetal acceleration. Would you just calculate it like:
Fn + mg = mv^2/R and solve for v?
if its like that how would you determine the normal force?
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Valentin is correct, almost.In the first paragraph, there is a slip: Here is his paragraph corrected:
At the top of the trajectory there are only 2 forces on Ș the centrifugal force directed inward the curve and the weight directed toward the center of the curve. The minimum speed is reached when these 2 forces are equal.
At the top of the trajectory there are only 2 forces on Ș the centrifugal force directed inward the curve and the weight directed toward the center of the curve. The minimum speed is reached when these 2 forces are equal.
Please excuse but the centrifugal force is BY DEFINITION directed OUTWARD the curve!!!
I am a PhD in physics
I am a PhD in physics
I missed your wording,you did make another critical error.
Here it is corrected:
At the top of the trajectory there are only 2 forces on Ș the centripetal force directed inward the curve and the weight directed toward the center of the curve. The minimum speed is reached when these 2 forces are equal.
I should have caught that error earlier.
Thanks for pointing it out.
Here it is corrected:
At the top of the trajectory there are only 2 forces on Ș the centripetal force directed inward the curve and the weight directed toward the center of the curve. The minimum speed is reached when these 2 forces are equal.
I should have caught that error earlier.
Thanks for pointing it out.
In case of any of you students did not catch the error, here is a good explanation of the error of the student claiming to be a PhD..
http://www.newton.dep.anl.gov/askasci/phy00/phy00305.htm
We shouldn't claim that which is not true, especially at Christmas, when St Nick is checking twice if we have been naughty or nice.
http://www.newton.dep.anl.gov/askasci/phy00/phy00305.htm
We shouldn't claim that which is not true, especially at Christmas, when St Nick is checking twice if we have been naughty or nice.
In teaching physics and engineering in the United States, the use of the term "centrifugal force" is usually discouraged, since it is a pseudo-force (or virtual force) that one must add when using a frame of reference that is rotating. If you use that approach, which is equivalent to d'Alembert's principle, you can use a rotating or accelerating frame of reference, with the virtual force opposite to the direction of acceleration, and treat it as a problem in statics, as Valentin has done.
If one uses an inertial frame of reference, no centrifugal force is introduced. Gravity and the normal force of the tracks provide the centripetal acceleration, which is inward. At minimum velocity for staying on tracks, there is zero normal force from the tracks.
If one uses an inertial frame of reference, no centrifugal force is introduced. Gravity and the normal force of the tracks provide the centripetal acceleration, which is inward. At minimum velocity for staying on tracks, there is zero normal force from the tracks.
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