Senior 1 physics problem uniform circular motion .

Assuming that the radius of the arch bridge r, the mass of the car m, and the velocity v then according to the first case 100m r=mg-n=mg 4....1) If the car is not subjected to frictional force, then f= n=0 =>n=0 so the centripetal force is completely provided by gravity v 2m r=mg....2) by (1)(2) =>v=20m s

If you don't test a uniform circular motion in the college entrance examination, it doesn't matter if you can't figure it out.

On the first floor, how do you know that you live in the same province as the building.

No friction, i.e. the bridge is not under pressure, and the centripetal force is provided entirely by gravity.

"When the speed of the car passing through the apex of the arch bridge is 10m s, the pressure of the car on the top of the bridge is 3 4 of the weight of the car".

m(v1^2)/r=mg-3/4mg

Find r=40m

By "the centripetal force is provided entirely by gravity".

m（v2^2）/r=mg

v2=20m s

Pulling each other, the rope provides the centripetal force, so the centripetal force is the same, and the root turntable rotates together, so the angular velocity is the same.

by f=m squared r

The radius ratio is equal to the inverse mass ratio of 2:3

Linear velocity v= r

The ratio of linear velocity is equal to the radius ratio of 2:3

BCD: When the string touches the nail, the radius is halved, and the center of the circular motion becomes point C, because the rope only touches the nail, the ball does not change the velocity by other external forces, that is, the linear velocity of the ball remains unchanged, and w becomes larger from w=v r; From a=v 2 r, it can be seen that a increases; When passing the lowest point, the centripetal force f-mg=mv 2 r, and f=mg+mv 2 r is obtained, and f can be judged to increase, so bcd is chosen

Excerpt from "Complete Interpretation".

When the BCD overhang is directly below, the velocity is constant, so the linear velocity remains unchanged.

The linear velocity does not change, the radius decreases, the angular velocity increases, the centripetal acceleration increases, and the required centripetal force increases. The centripetal force is the resultant force of gravity and the pulling force of the rope, so the pulling force increases.

Note that in this formula, to compare the relationship between two physical quantities, we must first ensure that the other quantities in the formula remain unchanged.

The frictional force provides the centripetal force and is mrw 2=10n

The greater the r, the greater the centripetal force required.

Of course, the farther away the easier it is to slide.

The maximum static friction is 1N

From the first question formula, w takes 10 rad s below the root, just without moving.

Therefore, the angular velocity is less than or equal to this value.

The answer is yes.

Let the angle be , the force on the ball is analyzed by gravity and elasticity, and the resultant force direction of the two forces is horizontally directed to the vertical axis of the ring, and the centripetal force formula is obtained from the formula of centripetal force to m* 2*(rsin) and f to mg tan

Two-way Polaroid cos g (r*2) 10 [ , so 60 degrees.

The analysis can be leaked, only A rope is stressed, and W is the smallest; Only the B rope is stressed, and W is the largest.

The combined force of gravity and the large tensile force of the collapsed pure rope provides the centripetal force.

1. Only a is forced.

3／3mg＝mrw^2

r=1 (this can be seen from the empty shirt of the figure).

w=2, only b is forced.

mg＝mrw^2

w＝3．16rad/s

In summary: (rad s).

1l The first block guessed that the second question was not calculated correctly.

It should be simple lead: let their masses be m, the angular velocity is w, and oa=r, then for oa to Zhaolu: f1=2m*w 2*3 2r=3m*w 2*r, for ab, f2=m*w 2*r=2m*w 2*r, so the ratio of tensile force is 3:2

1) Let the initial ball speed be w, and the initial tension of the rolling thin line is f.

The tensile force acts as a centripetal force according to the fine wire, there are:

mrw^2=f

mr(3w)^2=f+40

Solve the above two equations together, and the initial f is obtained.

2) w is solved by 1) the remaining two formulas in the auction.

Because the ball always moves in a uniform circular motion, there is v=3wr. (3w is the second speed, and the original formula should be equal to the speed of the ball multiplied by the length of the rope.) ）

Distance between the ball and the table: d=ssin60°

Note that the distance is d, not s.

Turn left|Turn right.

The resultant force of gravity and the pull of the rope acts as the centripetal force of the ball.

It can be easily solved by drawing a force analysis diagram.

In a right-angled triangle, it is not surprising that the right-angled side is smaller than the hypotenuse of several trillion tombs, and the tensile force is less than the gravitational force.

The B end is fixed on the rotating shaft of the turntable, then the thin rod is rotating, then, the situation sought must be the result of a stable state, that is, at this time the ball is also moving in a uniform circular motion, and its rotational angular velocity is 5rad s, you use force balance to solve, the tension of the string is decomposed into two directions: vertical and horizontal, the vertical part cancels the gravity, and the horizontal part is the centripetal force.

Linear velocity definition: describes the speed of the movement of the particle along the circle, is the vector, and the size of the linear velocity is the ratio of the arc length s of the motion of the particle to the time t used.

Definition of angular velocity: The time it takes for an object moving in a uniform circular motion to turn around in a circle. The magnitude is the ratio of the radian of the radius to the time taken in a certain amount of time.

Centripetal acceleration: When an object moves in a circle, the external force pointing to the center of the circle along the radius (or the component force of the external force pointing to the center of the circle along the radius) is called the centripetal force, and the acceleration generated by the centripetal force is centripetal acceleration. Centripetal acceleration is used to describe how fast or slow an object changes in the direction of linear velocity as it moves around a circle.

In order for m to be stationary, the upper limit of angular velocity is when.

When the gravity of m and the maximum static friction force together disperse the centripetal force, the maximum static friction force is directed along the radius to the center of the circle, that is, the hole, and its function is not to let the ball fly out because of the speed. At this time, if the W is a little bigger, then the whole system will do centrifugal motion, and the ball will fly out.

Cause. 2+mg=mw*wr

Find the upper limit of w.

Note that the lower limit of the method should be compared with the gravity of the maximum static friction of 2n and m to be smaller than the gravity force of 3n, which means that if the angular velocity of the ball is 0, when it does not rotate, the ball will be pulled down, so it must be pulled down to make it not be pulled down, when the rotation speed is very slow, that is, when mw*wr is less than mg, the excess part of mg-mw*wr wants to pull the ball down, so at this time, the friction force points outward along the radius, hindering the rush of excess gravity, but the maximum friction can only be provided to 2n, Therefore, when mg-mw*wr=2n, the angular velocity is the minimum angular velocity, and if it is smaller, that is, when mg-mw*wr>2n, the excess part of gravity is more than 2n, and the ball is pulled down.

So order. mg-mw*wr=2

Find the lower limit of w.