The illustration shows a version of the loop-the-loop trick for a tiny auto. If the auto is offered an initial rate of 4.0 m/s, what is the biggest worth that the radius r deserve to have if the vehicle is to remajor in call via the circular track at all times?

Drawing: \$\$P_gravityf = mg(2r)\$\$\$\$K_i = mg(2r)\$\$

However, to usage the theorem of conservation of power I was lacking one last component, the kinetic power at the peak, so I started doing this:

\$\$F_centripetal = fracmV^2r = mg + N\$\$

Now this is wright here I acquired stuck, however was later told by the explanation that centripetal force equals this:

\$\$F_centripetal = fracmV^2r = mg\$\$

So I am guessing they assumed normal force will certainly equal to 0. However, exactly how is that possible? Wouldn"t the automobile lose contact if the normal pressure was 0?

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edited Feb 22 "17 at 22:27 Qmechanic♦
asked Feb 22 "17 at 21:53 PabloPablo
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So a constraint force, and the normal pressure is an instance of this, enpressures some predetermined equation \$f(x, y) = 0.\$ The constraint which says "the auto must remain on this circle of radius \$r\$" is provided (in coordinates wright here the origin is at the facility of the circle) as \$x^2 + y^2 - r^2 = 0,\$ for instance.

The constraint pressure will in general be yet solid it needs to be to enpressure the constraint, and it will allude along a direction \$ abla f = mathbf i ~ fracpartial fpartial x + mathbf j ~ fracpartial fpartial y\$, additionally recognized as the "gradient" of \$f\$. This is bereason the gradient happens to allude in the direction of biggest increase, and that happens to be perpendicular to this "level set" \$f =0.\$ On this account I am providing you, the constraint force deserve to suggest in the direction of either \$+ abla f\$ or \$- abla f,\$ whatever it needs to store the ppost on the track \$f(x, y) = 0.\$

Now your track is a tiny one-of-a-kind, and also different from what I simply shelp, bereason the car is not permitted to go through the track, yet is allowed to fall off it. In various other words, the constraint is \$f(x, y) le 0\$ quite than \$f(x, y) = 0.\$ The just change that you need for this case is "the constraint pressure just exists as soon as \$f(x,y) = 0 \$ and also only if it points in the negative direction \$- abla f,\$ if it instead were to suggest along the positive direction \$+ abla f\$ then that means that the object normally is trying to leave the constraint by decreasing \$f\$ which it is completely enabled to do: no constraint pressure needed."

So the closest that you have the right to obtain to falling off of the track without actually falling off, needs that you modify these other parameters till the minimum constraint pressure viewpoints 0. Then if you made the track any larger, the constraint force would start pointing in the direction of \$+ abla f\$ and therefore you would certainly be falling off the track, by the above reasoning. So that is why the normal force has to go to 0 at the optimal at the important radius: if the radius acquired any kind of larger, then the constraint "I have to stay on this circle" would certainly generate a force pulling the auto towards the loop, but the loop cannot "grab" the auto and "pull it towards the loop" without, say, the automobile having a component wrapped roughly a railing a lot like a roller-coaster "holds on" to its track. Just tires on a surface will not pull the automobile in the direction of the surchallenge if it is falling off.

If you want additionally confirmation, you deserve to execute this trouble a various way: imagine we simply begin through the auto at the optimal of the loop, \$(x, y) = (0, r)\$, firing it with a rate \$v\$ in the \$-x\$ direction. We know such a things in free-autumn describes a parabola \$x(t) = x_0 - v t,~~y(t) = y_0 - frac 12 g t^2.\$ However we additionally recognize that the curve \$y = sqrtr^2 - x^2 = rsqrt1 - (x/r)^2 approx rleft(1 - frac 12 left(frac x r ight)^2 ight) = r - fracx^22r.\$ Using \$x_0 = 0, y_0 = r\$, the first equation gives \$y = r - frac12 g (x/v)^2,\$ and also requiring it to be a tangent parabola to the circle therefore provides \$g/v^2 = 1/r.\$

If \$v\$ were any smaller sized, the parabola would certainly be too narrowhead and it would certainly "fall off" the circle; however \$v\$ can of course be bigger and the constraint pressure will certainly "save it on" the circle. Then you think around this instrumental case and you realize, "oh, in this case the approximate circular movement is gave totally by free-autumn and therefore gravity is doing every little thing I should be in circular movement, the normal pressure should therefore be zero."