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Understanding the Minimum Speed to Keep Carriage on Tracks in a Loop
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[QUOTE="User1265, post: 6308929, member: 667792"] [B]Homework Statement:[/B] Carriage of rollercoaser about to enter a vertical loop of diameter 17m. The carriage is initially at rest delta h above the bottom of the loop. Such that the passenger remains in contact with their seat, calculate the minimum speed of the car at the top of the loop. [B]Relevant Equations:[/B] mg = mv^2/R I recognise that the normal force must alwayss act towards the centre of the circle loop, as the rail always has to be exertign a pushing force on the car/carriage in order for it to follow the trajectoryof the loop. However , I cannot understand why, the reaction force has to be greater than or equal to zero in Q1 and Q2. ----------------------------------------------------------------- My solution : F net = mv[SUP]2[/SUP]/R = F[SUB]normal[/SUB] + mg where m - mass of whole carriage (inc. seat + passenger) then F[SUB]normal[/SUB] greater than or equal to 0 --> for, mv[SUP]2[/SUP]/R to be greater than mg , to allow a weight of value mg going round the top of the tracks. Otheriwse If mv^2/R <mg - hoz velocity at the top of track will be too small to continue on the track and the carriage will fall off initally a projectile manner --------------------------------------------------------------------------------------------- I understand that for the whole carriage ( including the passenger and seat) to stay on the track, then the centripetal force in the limiting case of minumum speed, would have to be at least equal to the weight of the whole carriage in order to produce a great enough speed that will keep the trajectory (of that weight) at that instant - thus staying on the tracks at that instance of the carriage at the top of the loop. But this is assuming the only force acting in radial plane is weight , as normal force is assumed to be zero in the limiting case. But Q1) is why can the normal force be assumed to be zero in the case of slowest speed to keep a circular trajectory? With a zero Normal force, would passenger and seat not be in physical contact? So wouldn't the minimum speed really need to be slightly greater than Fn =0 such to keep physical contact between seat and passenger, or is it that lack of physical contact is negligble , and thus even with Fn=0 can the smallests safe, velocity be produced in reality. Q2) Also, Surely if the seat exerted an upwards force on the passenger, to somehow stick onto the seat, isn't the condition that the passenger remains in contact with their seat is still satisified , regardless of whether the centripetal force (net force radially) is too small (<weight whole carriage) to produce a sufficient velocity to keep the whole carriage on the track at the top of the loop, and falling off. So surely the minium speed at the top can be zero, and the seat and the passenger can both fall at the same rate, where lack of physical contact is again neglible. [/QUOTE]
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Understanding the Minimum Speed to Keep Carriage on Tracks in a Loop
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