I was wondering this: Why is the $\sqrt{gR}=v$ the same for finding minimum speed and maximum speed? Like if you are wondering the min. velocity for keeping a rollercoaster on track when on top of the loop, you would use $\sqrt{gR}=v$ or like finding the Max speed without losing contact on the top of a hill.
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$\begingroup$ Have you drawn a free body diagram for each of these two situations? $\endgroup$– Chet MillerCommented Dec 17, 2018 at 13:51
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2 Answers
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At the top of the roller-coaster the equation of motion for the carriage, with down positive, is $mg+N=m\dfrac {v^2}{R}$ where $m$ is the mass of the carriage and $N$ the force on the carriage due to the rails.
The condition $v= \sqrt{gR}$ is satisfied when $N=0$.
If the speed is larger than this then $N$ is positive and so the rail must exert a downward force on the carriage to keep it on the rails.
If the speed is smaller than this then $N$ is negative and so the rail must exert a upward force on the carriage to keep it on the rails.
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$mg+F=m\frac{v^2}{r}$, where $F$ can be the tension in a ball and string system, or the normal force in the track and roller coaster system.
Minimum $v$ required for the object to make the loop is when $F=0$, and centripetal force would provide solely for the weight. That is, $g=\frac{v^2}{r}$, and solving gives $v=\sqrt{rg}$.
For the second part, it is actually like the first. In order to not lose contact with the hill, you are requiring centripetal force to provide for the weight such that it can undergo centripetal motion; too fast a speed will cause it to escape. Derivation will be same as above.
In both scenarios, notice the words "provide" and "require". Both problems essentially are the same. We are finding the condition that the object can undergo centripetal motion; which is when the centripetal force and provide for what is required by the weight to balance it out.
answered Dec 17, 2018 at 13:44