The illustration shows a version of the loop-the-loop trick for a small car. If the auto is provided an initial speed of 4.0 m/s, what is the largest value that the radius r can have if the automobile is to continue to be in contact with the circular monitor at every times?
I made two assumptions initially:
$$P_gravityf = mg(2r)$$$$K_i = mg(2r)$$
However, to use the organize of conservation of power I was lacking one last component, the kinetic energy at the top, for this reason I began doing this:
$$F_centripetal = \\fracmV^2r = mg + N$$
Now this is where I acquired stuck, yet was later on told by the explanation that centripetal force amounts to this:
$$F_centripetal = \\fracmV^2r = mg$$
So i am guessing castle assumed normal force will equal to 0. However, exactly how is that possible? Wouldn\"t the automobile lose call if the normal force was 0?
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edited Feb 22 \"17 at 22:27
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So a constraint force, and the normal force is an instance of this, enforces part predetermined equation $f(x, y) = 0.$ The constraint which claims \"the vehicle must continue to be on this one of radius $r$\" is given (in collaborates where the origin is in ~ the facility of the circle) together $x^2 + y^2 - r^2 = 0,$ for example.
The constraint pressure will in basic be however solid it needs to it is in to enforce the constraint, and also it will point along a direction $\\nabla f = \\mathbf ns ~ \\frac\\partial f\\partial x + \\mathbf j ~ \\frac\\partial f\\partial y$, additionally known as the \"gradient\" that $f$. This is because the gradient wake up to suggest in the direction of biggest increase, and also that wake up to be perpendicular to this \"level set\" $f =0.$ top top this account ns am giving you, the constraint force can suggest towards either $+\\nabla f$ or $-\\nabla f,$ whatever it needs to save the fragment on the track $f(x, y) = 0.$
Now her track is a tiny special, and different from what I simply said, since the automobile is not allowed to go through the track, however is permitted to fall off it. In other words, the constraint is $f(x, y) \\le 0$ fairly than $f(x, y) = 0.$ The only alteration that you require for this situation is \"the constraint pressure only exists once $f(x,y) = 0 $ and also only if the points in the negative direction $-\\nabla f,$ if it rather were to allude along the optimistic direction $+\\nabla f$ climate that way that the object naturally is trying to leave the constraint by to decrease $f$ which the is totally allowed to do: no constraint force needed.\"
So the closest the you can acquire to falling off of the track without in reality falling off, needs that you modify these other parameters till the minimum constraint force approaches 0. Then if you made the track any kind of larger, the constraint pressure would begin pointing towards $+\\nabla f$ and also therefore you would be falling off the track, through the over reasoning. So the is why the normal force has to walk to 0 in ~ the top at the an important radius: if the radius got any larger, then the constraint \"I should stay top top this circle\" would generate a force pulling the vehicle towards the loop, however the loop cannot \"grab\" the car and \"pull it in the direction of the loop\" without, say, the car having a component wrapped about a railing much like a roller-coaster \"holds on\" to its track. Simply tires top top a surface won\"t traction the auto towards the surface if it is fall off.
If friend want more confirmation, you can do this trouble a different way: imagine we just start with the vehicle at the top of the loop, $(x, y) = (0, r)$, firing it through a rate $v$ in the $-x$ direction. We understand such an object in free-fall explains a parabola $x(t) = x_0 - v t,~~y(t) = y_0 - \\frac 12 g t^2.$ however we likewise know that the curve $y = \\sqrtr^2 - x^2 = r\\sqrt1 - (x/r)^2 \\approx r\\left(1 - \\frac 12 \\left(\\frac x r\\right)^2\\right) = r - \\fracx^22r.$ utilizing $x_0 = 0, y_0 = r$, the first equation provides $y = r - \\frac12 g (x/v)^2,$ and also requiring the to it is in a tangent parabola to the circle as such gives $g/v^2 = 1/r.$
If $v$ were any smaller, the parabola would certainly be as well narrow and it would certainly \"fall off\" the circle; however $v$ deserve to of course be larger and also the constraint force will \"keep that on\" the circle. Climate you think about this an essential case and you realize, \"oh, in this case the almost right circular activity is provided entirely by free-fall and therefore gravity is doing every little thing I must be in circular motion, the normal force must because of this be zero.\"