Problems of ball mechanics, theoretical mechanics, horizontal ball rolling

Let the support force of each ball at the bottom to the fourth ball be n, the radius of the ball is r, the radius of the bowl is r, the support force of the bowl to each ball is n', and the force between the three balls at the bottom is f. First, the force analysis of the fourth ball is carried out, and the force of gravity is set to mg and the net force of the three balls at the bottom to support it. Four balls form a tetrahedron (flute length is 2r), and the resultant force of the supporting force can be found to be 6n, and the force equilibrium can be known as 6n=mg.

Then take one of the three balls at the bottom to perform a force analysis, gravity, the bowl support force n', the pressure of the fourth ball on it -n, and the net force f of the pressure of the other two balls on it (can be obtained by force analysis), respectively, the bowl support force n', the pressure of the fourth ball on it -n is decomposed in the vertical direction, and the horizontal plane formed by the bottom three balls. The pressure of the fourth ball on it -n can be decomposed vertically by 1 3 mg (6 3N) and horizontally by 2 6 mg (3 3N) bowl support force n' can be decomposed into vertical n'( (r-r) 2-(r 2)4 3) r and n horizontally'(2r 3 3) r is balanced according to the force, in the vertical direction 1 3mg+mg=n'( (r-r) 2-(r 2)4 3) r in the horizontal direction f+ 2 6mg=n'(2r, 3, 3) r, and because f>=0, 2 6mg<=n'(2r 3 3) r According to the above two equations, the relationship between r and r can be obtained. r<=2 11r+rIn addition, when the radius of the bottom three balls is large, and the plane of its composition coincides with the mouth of the bowl, then r=(2 3+3)r 3, so r must be greater than (2 3+3)r 3, so (2 3+3)r 3=some calculations are omitted, if you want to know, you can contact me.

Hemispherical bowl, thickness is not counted""Can it be understood that the weight is not counted?? I studied mechanics, and it doesn't look like a problem that can be solved on the Internet. It takes time to analyze, pen and draft head.

The answer is wrong. Here's the answer:

The force exerted on the ball in the vertical direction is: p,fn; The force on the ball in the horizontal direction is: f,ff; The frictional resistance of the ball in the opposite direction of rotation is: MF.

The rigid condition of the ball rotating at a constant speed is: mf=(delta)*fn, ff<=(fs)*fn (1).

where delta is the rolling resistance coefficient and fs is the sliding friction factor.

From the equilibrium equation in the horizontal direction, f=ff (2).

From the equilibrium equation in the vertical direction, fn=p (3).

From the equation of moment equilibrium for the center of mass, mf=(ff)*r (4).

From Eq. (1), (2), (3), (4), f=(delta)*p r (5).

Eq. (1) becomes, ff=(delta)*p r, ff<=(fs)*p (6).

Discussion: If (delta) r<=fs, i.e., delta<=(fs)*r, Eq. (6) holds, f=(delta)*p r can make the ball roll at a constant speed.

If (delta) r>fs, i.e., delta>(fs)*r, Eq. (6) is not true, and in the process of increasing F, MF has not yet reached its maximum value (it is impossible for the ball to roll without reaching this value), and the static sliding friction has reached its maximum value, and the ball begins to slide without rolling.

Conclusion: Only when delta<=(fs)*r, f=(delta)*p r, can the ball roll at a uniform speed.