The two small balls AB with a mass of m are connected by a light rod with a length of 3 liters

Let the angular velocity of the pole be:

Then there is: va=2l, vb=l

Conservation by energy: 2mgl=mgl+m(va2+vb2)22mgl=ml2(4 2+2).

5ω^2=2g/l

2=2g/（5l)

2g/（5l)

va=2√2gl/5（m/s）

The force of ball A on the pole: f = f centripetal force + mg = m 2 * 2l + mg = 4 mg 5 + mg =

The direction is vertically downward. Size.

The force of the ball b on the pole: f=f centripetal force-mg=m2*l-mg=, the direction is vertically downward, and the magnitude is magnitude.

Then there is: the force of the rod on the rotating shaft o:

The direction is perpendicular downward, here it is the force of the rod on the shaft, the direction is downward, the answer says that the direction is up is wrong, if it is the force of the shaft on the rod, the direction is vertically upward).

va=2vb

Total E = E = E + E = 1 2m (VA + VB) = δe weight = 2 mgl - mgl = mgl

Vb = 2 5GL, VA = 8 5GLF centripetal A = MVA 2L = 4 5mg

F centripetal b = MVB l = 2 5 mg

The force analysis of the O is vertically downward in the direction of A:Fa=F, centripetal A+GA=4, 5mg+mg=9, 5mg.

For b: FB=GB-F centripetal B=MG-2 5MG=3 5mg direction vertically downward.

fo=fa+fb=7 5mg direction straight downward.

a c

A-ball: Gravity does negative work, the gravitational potential energy increases, and the kinetic energy increases.

B-ball: Gravity does positive work, the gravitational potential energy decreases, and the kinetic energy also increases, and only gravity does the work in the whole process, and the mechanical energy of the system is conserved.

Since the center of rotation o is the midpoint, the velocity of the two balls is the same, let the velocity be v, gravity does positive work on b, and negative work on a.

So 2mg(l2)-mg(l2)=1 2*mv squared + 1 2*(2m)v squared.

v = root number gl

Let the tensile force ff-2mg=(2mv squared) (l 2).

f=6mg

The B ball is subjected to 3 forces, the support force of the rod, along the direction of the rod, f, and gravity, so when the velocity is maximum, the 3 forces are balanced, and the resultant force is 0, so when the resultant force of f and g = the supporting force of the rod, that is, the angle of the rod is the opposite direction of the resultant force formed by f and gravity. At this time, the angle of turning =

According to the same angular velocity, it can be seen that the linear velocity of a is half that of b, and the work done by f = mg [2 (root number 5)].

Let the velocity of a be v, and the conservation of energy yields mg [2 (root number 5)] = and solution 2v =

There is a system for the two balls and the rod, obviously the mechanical energy of the system is conserved, when the rod reaches the vertical position, the velocity of ball A is v1, and the velocity of ball B is v2, notice that their angular velocity is the same, so there is v2 2*v1

From the conservation of mechanical energy, (m*v1 2 2) (2m*v2 2 2) 2m*g*(2l 3) m*g*(l 3).

Get v1 [ root number (2gl)] 3 , v2 2*[ root number (2gl)] 3

The centripetal force formula mg f1 m*v1 2 l 3) is used at the highest point of the ball A (since the direction of the force f1 of the rod on a is not known, here it is assumed that the force f1 of the rod on the ball A is upward).

Mg F1 m* 2 l 3).

f1 mg 3, which is a positive value, indicating that the direction of the force exerted by the rod on the ball A is indeed upward.

b ball at the lowest point, using the centripetal force formula f2 2m*g 2m*v2 2 2l 3).

f2－2m*g＝2m*^2 / 2l / 3)

The force of the rod on the b-ball is f2 14 mg3 and the force of the rod on the b-ball is upward.

From Newton's third law, the force of A on the rod is downward, and the force on B on the rod is downward, so the force on the rod at point o is upward.

The force of the rod to the point o is downward and its magnitude is f f1 f2 (mg 3) (14 mg 3) 5mg

Take the horizontal line where the point o is located as 0 potential energy.

1) According to the conservation of energy, the kinetic energy of two balls can be obtained, and since Bo=2AO, the velocity of b is twice the velocity of A, from which the velocity of B can be calculated.

2) Strive to draw out the centripetal force of A and B respectively (pay attention to the direction), and the gravity of A and B can be added to obtain the force and direction of point O.

AC analysis: in the process of releasing the light rod from the horizontal position from the stationary position to the vertical position, the system only has gravity to do the work, the mechanical energy is conserved, the downward movement speed of b increases, and the height decreases, a according to the definition of kinetic energy and potential energy, the change of kinetic energy potential energy can be judged, and the work done by the object according to the force other than gravity is equal to the change of the mechanical energy of the object to judge the work done by the rod on the ball a.

A and b increase the downward movement speed, so the kinetic energy increases, the height decreases, so the potential energy decreases, so a is correct;

b. A increases the speed of upward movement and increases the height, so the kinetic energy and potential energy both increase, so B is wrong;

The system composed of C, A ball and B ball has no other form of energy conversion in motion except kinetic energy and potential energy, so the mechanical energy of the system is conserved system, so C is correct;

D, A ball and B ball system only gravity does the work, and the mechanical energy is conserved, so D is incorrect;

So choose AC. Comments: This question is a model problem of light rod connection, the mechanical energy of the system is conserved, but the mechanical energy of a single ball is not conserved, and the work done by the object by using the conservation of the mechanical energy of the system and the force other than gravity is equal to the change of the mechanical energy of the object.

Solution: In this process, the B ball with a mass of 2m lands L 2, so the rod does negative work on the B ball, and the magnitude is: 2mgl 2= mgl

Then the rod does positive work on the A ball, and the magnitude is: MGL

mgl／2 +1／2mv^2=2mgl／2-（1／2*2*mv^2

v = gl 3 under the root number

All is the meaning of dividing.

(1) When A is at the highest point, there is mg=m for A and TOB-2mg=2m for B, and TOB=4mg can be obtained.

According to Newton's third law, the force on the O axis is 4mg, and when the direction is vertically downward (2)b at the highest point, there is 2mg+ t ob=2m for b, and t ob=0 can be obtained by substituting v in (1).

For A, there is T oa-mg=m, t oa=2mg.

According to Newton's third law, the magnitude of the force on the O-axis is 2mg, and the direction is vertically downward (3) In order for the O-axis to be not forced, the mass of B is greater than the mass of A, and it can be judged that the B ball should be at the highest point. There is T ob+2mg=2m for b and T oa-mg=m for A. When the axis O is not stressed, T Oa = T ob, v = 3gl can be obtained

According to the problem of the force on the o-axis, look at Newton's third law. Explained in detail.

For the ball aga+ma*w (angular velocity)*1 3*l=t, w=6m s

Centripetal force to the pellet bf = mb * w * 2 3l =

and GB=6N

Therefore, the light club of team B acts as a pulling force f=fcentripetal-gb= direction, straight upward.