Is it true that a circle is only smooth when it is timely? 50

Yes. Because when it is not smooth, there is friction.

Factors: The vertical chord friction is the smallest, and the greater the angle with the vertical direction, the greater the friction, and the complication of time and distance (displacement). - Of course, there are cases of inclined planes, but they are not isochronous circles.

characteristics.

I hope you can accept it, it's not easy to be serious. Your question should be a physics question: What is the time phase in which an object slides from rest along all the smooth strings that pass through the highest point on the same vertical circle to the lowest point of the circumference?

Answer: Since each string is smooth, the object slides downwards, now prove the property of time in the process of a certain string sliding downward. From the linear motion of uniform acceleration, 2rcosa = at(square) 2, acceleration a = mgcosa m = gcosa, the two equations get t=radial sign 2r g, we can see that t falls along the diameter when it is only related to r.

r is the radius and a is the angle between the diameter and the string) This proves that the time is the same no matter which string it falls on, and is called an isochronous circle. Please accept. It's not easy.

Isochronous circle: Let a circle o radius r, a is the highest point of the circle o, b is any point on the circle, an object starts from a and slides down ab to b, the time taken is equal, it is the time taken from a free fall to the lowest point of the circle.

The condition of a physical isochronous circle is: orbital smoothness a=gcos The angle between the orbital and the vertical direction.

ab=2rcosθ=1/2gcosθ t^2 t=2(r/g)^1/2

Smooth. Only when the friction is negligible, there is isochronism.

Isochronous circle. Let a circle O radius r, A is the highest point of the circle O, and B is the point of any clear disturbance on the circle, and the time taken for an object to start from A and slide down Ab to B is equal, and it takes the same amount of time to go from A to the lowest point of the circle.

Physically, the bar of an isochron circle is: the orbit is smooth a=gcos The angle between the orbit and the vertical direction.

ab=2rcosθ=1/2gcosθ t^2 t=2(r/g)^1/2

This is not an equivalence circle. There are only two types of isochronous circles.

<> only these two cases can use the isochronous circle model, and the other cases are not.

Equivalence Circle Certificate:

Take a as an example, toa = [2rsin ado) (gsin ado)] = 2r g).

The result is independent of the angle, and the isochronous circle conclusion holds.

About your **, in fact, as long as you list similar formulas according to the idea of deriving isochronous circles, it can be solved, in the final analysis, it is still to set angle variables and list functions to solve the problem. (Equivalence is not necessarily equivalence).

How to crack the mutated version of the isochronous circle.

Can you make the problem clear?

It takes the same amount of time to slide down a smooth straight track to an arc from the highest point of a circle, or from a point above a circle along a smooth straight track to the lowest point.

Let a circle O, A be the highest point of the circle O, and X is any point on the circle, and the time taken for an object to start from A and slide down ax to X is equal, which is the time taken from A to the lowest point of the circle.

Proof: Since each string is smooth and the object slides along, the characteristics of time in the process of sliding along a certain string are demonstrated.

From the uniform acceleration of linear motion, 2rcosa = at (squared) 2, and the acceleration a = mgcosa m = gcosa, the two equations are obtained.

t = root number 2r g, knowing that t is only related to r when falling along the diameter. (r is the radius, a is the angle between the diameter and the chord).

This proves that no matter which string falls, the time is the same, which is called an isochronous circle.

Derivation steps for physical isochronous circles:

Connect the highest and lowest points of the circle, and connect the lowest point to any point on the circumference according to x=1 2*a*t 22r=1 2*g*t 2 t=2 (r g), assuming that the angle is a, then the length of the inclined plane is 2rcosa and the acceleration is a=gcosa

According to x=1 2*a*t 2

2rcosa=1/2*gcosa*t^2

t=2√(r/g)

Let the angle between the smooth chord and the vertical direction be , and the radius of the circle is r, then the acceleration a=gcos, the displacement s=2rcos, and the object moves in a uniform accelerated linear motion with a muzzle velocity of 0, s=1 2at 2, and t=root number (4r g) are equal.

This is very simple, if the racer does not know that the "isochronous circle" can be pushed out with an extreme value, this method is less useful and more flattering.

An isochronistic circle is a circle in which a smooth string slides freely from rest to another point in the circle (uniform acceleration) from the highest point in the same circumference.

Take your diagram as an example, if the radius is r, then ab=2rcos, and the acceleration along ab is gcos, apply the formula for guessing the velocity of the uniform band with initial velocity of 0, t = root term (2ab a) = root term (2 2rcos gcos) root term (4r g).

So which way is the smallest, the fastest, as long as you are satisfied with the intersection of the circle and the inclined plane (that is, the point on the circumference).

This is very simple, if the racer does not know that the "isochronous circle" can be pushed out with an extreme value, this method is less useful and more flattering.

An isochronistic circle is one in which a smooth string slides freely from rest to another point on the circumference (uniform acceleration) from the highest point of the same circle, starting from a standstill.

Take your diagram as an example, if the radius is r, then ab=2rcos, and the acceleration along ab is gcos, apply the formula of uniform acceleration motion with initial velocity of 0, t = root term (2ab a) = root term (2 2rcos gcos) = root term (4r g).

So whichever garden is smaller is the fastest, as long as you have an intersection point between the circle and the inclined plane (i.e., a point on the circumference).

Let the diameter of the circle in the vertical plane be d, then ac=l=dcos The time taken for the particle to slide down from rest along the smooth inclined plane ac is t, and the acceleration of the particle is a=gcos (Newton's second law), according to the displacement formula l=(1 2)*a*t 2

The physical meaning of >>the result is that the time sought is exactly the time taken by the particle to move from A to B in free fall. It has nothing to do with the inclination of the inclined plane, so we get a conclusion: from the highest point of the circle to do the secant, the movement time along the secant is equal, so it is called an isochronous circle.

The secant released from any point on the circle to the lowest point of the mass is released at rest, taking the same amount of time and equal to the time it takes to make a free fall along the vertical diameter.

That problem does not have to wait for the acceleration of the garden a: the acceleration of the small distance is large, the acceleration of b is medium, the distance is medium, and the distance c is the shortest, and the acceleration is the largest.