A light rope across a fixed horizontal smooth thin rod O, with a small ball tied at each end...

It's a difficult question. The first question is the idea: A just left the ground to indicate that the pull force of the rope is mg The rest is a mathematical problem--- the vertical downward gravity of the ball A and the tension of the rope mg The resultant force direction is the direction of the outer tangent of the arc Therefore, the angle is 0, that is, when b moves to the lowest point (I am also surprised to get this result, after all, the high school physics knowledge is almost forgotten, and I vaguely feel that something is not right, the result may not be correct, and I look forward to the answer from the master.)

The second question idea: let the angle be the velocity v0 of the ball at this time can be found (at this time, the distance of the ball from the initial unknown height can be found, and the gravitational potential energy can be converted into kinetic energy to find the velocity v0) v0 direction is the direction of the vertical string - that is, the tangent direction of the arc, the velocity is decomposed, and the expression of the vertical velocity v1 is found to solve the limit.

The third question should be a cube. Explanation: In the case of equal weight, the volume of the cylinder is smaller When pushing down, the connecting cylinder between the push point and the force point is shorter, so the amount of the center of gravity of the cylinder is raised less than that of the cube, so it is concluded.

1 Let the radius of the circular motion of the B ball be L, and the angle sought is the angle A.

mgl*cosa=1 2mv 2 kinetic energy theorem to find the velocity.

Mg-mgcosa = MV 2 l Centripetal force vs. velocity in the critical state.

I calculated a=arccos1 3

2 When the vertical component of the rope tension is equal to the gravitational force, the vertical velocity is the largest, and then the vertical acceleration is upward, which is still the first question.

mgl*cosa=1/2mv^2

f-mgcosa=mv^2/l

fcosa=mg

I calculated a = arccos root number 3 3

3 If the base area is the same, the side length of the square is larger, the cube should be more difficult to push down, the application is the knowledge of moment, the longer the side length, the longer the line of action from the fulcrum to the center of gravity, the greater the g*l2 resistance moment, and the power arm should be half as high (if the force is applied from the middle point), because fl1=gl2, the greater f.

Three good questions, don't forget to draw.

This can be translated into a leverage problem (this should be known), the support point of the leverage at point C.

The mass of A is 4 times that of B, and when stable, AOC and BOC are the same size, as the perpendicular line of A and B about Co, the two right triangles are similar, and the ratio of their perpendicular lines is inversely proportional to the mass, and the ratio of OB and OA can be obtained from similar triangles and the ratio of their perpendicular lines, and OB should also be 4 times that of OA.

My high school teacher taught me, and it's a great way.

The combination of physical and mathematical methods can be solved, you can analyze the force of the two balls of A and B respectively, the two balls are equal to the tension of the rope and the support force of the rod, and then combine the knowledge of the balance of the three forces, the resultant force of any two forces is equal to the third force, draw a composite diagram of the force, combine the dotted lines in the figure, so that the triangle of the force is similar to the geometric triangle, you can get the answer, your answer is the correct answer.

Classic topic, you draw a force diagram. Then the three forces are balanced by the force, so the three forces can form a triangle. The point of contact between the small ball and the large ball is called A.

Separated from the triangle and the triangle oo'a Similar.

Then correspond to the gravity support force of the pull force of the three sides with oo' oa o'a Three sides correspond.

So Pull: Support: Gravity = L: R: (H+R) I don't know if you know that.

A lot of these topics.

Connect the o points and m points, and the direction of the pulling force is o'm, the elastic direction is in OM and the gravitational direction is in OO', because the forces are balanced, the three form a closed triangle, so that the ratio of the three forces is equal to the triangle o'The ratio of the length of each side of om.

First of all, the ring is always immovable, so it is static friction, it has nothing to do with positive pressure to look at the pressure first, the ring and everything below it is regarded as a whole, the vertical direction is only the gravity and the support force of the rod (the force on the rope is the internal force, the integral method ignores it), the support force is always equal to gravity, the support force and pressure are the action force and the reaction force, so the pressure is always equal to the gravity, and the pressure is unchanged.

Looking at the friction again, the angle of the rope changes, so that the tensile force on the rope increases in the horizontal direction, the ring remains stationary, and the static friction force is equal to the horizontal component of the rope, so the static friction increases.

To sum up, choose B

Answer B takes the two rings as a whole, and the vertical direction is only affected by gravity and the supporting force, so the supporting force always remains the same. That is, the pressure remains the same.

Horizontally, only friction and tension are subjected to. The more inclined the rope, the greater the pulling force required. It can be seen that the friction force increases.

1) At the lowest point, f mg m*v 2 l is obtained by the centripetal force formula

The rope pull force is f mg m*v 2 l

2) (m*v 2 2) (m*v 1 2 2) mg*2 l is obtained from the conservation of mechanical energy

i.e. v 2 v1 2 4g*l .Equation 1

If the rope is broken at the lowest point, the ball does a flat throwing motion with muzzle velocity v, and rolls by s1 v*t1 and h g*t1 2 2 (h is the height from the lowest point to the ground), and the s1 v* root number (2*h g) is obtained

If the rope is broken at the highest point, the ball is a flat throwing motion known as v1 in muzzle velocity, which is composed of s2 v1*t2 and h 2*l g*t2 2 2 to obtain the s2 congratulatory segment v1* root number [ 2( h 2*l) g ].

By the question condition s1 s2, so v*root(2*hg) v1*root[2(h2*l)g].

i.e. v 2*h v1 2*(h 2*l).Equation 2

From Eq. 1 and 2 we get v 2*h (v 2 4g*l)*(h 2*l).

Finishing 2v 2 8g*l 4gh

So the distance sought is h (h l) (v 2 2g*l) 2g).

3) I didn't see the specific content of your question, the relationship between the height h of point o from the ground h and the length of the rope l should be satisfied is the result expression h (v 2 2g*l) 2g) in question 2).

1 As can be seen from the figure, in the process of flat throwing, the ball moves in the horizontal direction with a length of one-half of the root number three l, and in the vertical direction, there is a length of one-half of the length of the four-state regret ,......There is in the vertical direction gt = l; in the horizontal direction vt= 3·l ; muzzle velocity v0=v;

2. When the rope is just straightened, the speed of the ball is along the vertical direction of the rope, and then the rent is quickly reduced to 0; So, mgl=mv l; The solution yields v=......I'm not going to forget it, uh-huh, it's done.

(1) According to the centripetal force formula f-mg=mv

l Yes: f=mg+mvl;

2) When the ball moves to the lowest point, and the rope is suddenly disconnected, the ball is flat throwing for t, then h-l=12gt

x=vt let the velocity of motion to the highest point be v , which is obtained by the conservation of mechanical energy

2mgl+1

2mv′=12

After the MV ball moves to the highest point and the rope is disconnected, the ball is thrown flat for t, then h+l=1

2gt′x′=v′t′

and x=x

The above solutions are: h=v

2g-l3) the minimum value of centripetal force when the ball moves to the highest point is mg, then there is: mg mv l so v

gl then by the law of conservation of mechanical energy:

2mgl+1

2mv′=12

The velocity of the MV ball when it moves to its lowest point is v

5GL is therefore asked by (2) result h=v

2g-l≥5gl

2g-l=3

2L, i.e. H3

2l Certified

Answer: (1) When the ball passes through the lowest point, the size of the rope tension f on the ball is mg+mvl; (2) The height h of point o from the ground is v

2g-l；(3) Prove that the height of point O from the ground h and the length of the rope l should meet the requirements of h 32l see above