The radius r digs out the center of the circle with half R
Publish: 2021-03-24 07:39:38
1. If they are concentric, the center of gravity remains unchanged
if they are not concentric, let the center of gravity of small circle a and the center of gravity of large circle B connect AB and extend AB to C such that BC = AB / 3, and point C is the center of gravity of the rest
the idea is as follows: on the contrary, if the mass of the excavated circle is m, the mass of the remaining part is 3M, and the center of gravity of the combined circle should be on the line between the center of gravity of the small circle and the desired center of gravity, and the distance from the center of gravity of the small circle: the distance from the center of gravity of the remaining part is 3:1, and the combined center of gravity is the center of gravity of the original circular plate. So~~
if they are not concentric, let the center of gravity of small circle a and the center of gravity of large circle B connect AB and extend AB to C such that BC = AB / 3, and point C is the center of gravity of the rest
the idea is as follows: on the contrary, if the mass of the excavated circle is m, the mass of the remaining part is 3M, and the center of gravity of the combined circle should be on the line between the center of gravity of the small circle and the desired center of gravity, and the distance from the center of gravity of the small circle: the distance from the center of gravity of the remaining part is 3:1, and the combined center of gravity is the center of gravity of the original circular plate. So~~
2. The center of gravity is on the opposite extension line of the line connecting the two centers
3. Analysis 1: if the mass of the ball with radius R / 2 is m ', then M' = the gravitational force of the remaining part on the ball m which is d away from the center of the ball f =. Analysis 2: the mass of the ball with radius R / 2 is, and the mass of the remaining part is. Since the remaining part is a spherical shell with uniform mass distribution, it can be considered as a particle with mass concentrated in the center of the ball, According to the law of gravitation, f ==
4. It should not be difficult to calculate the moment of inertia of a small circle by using the difference method or finding the center of gravity first and then integrating
5.
选择D
6. Take the part of the small ball as negatively charged (the charge density is equal to that of the large ball), and solve the equation (7 / 8) ×( four π/ 3) × r^3* ρ＝ Q get ρ Then the big ball can be regarded as a whole uniformly charged with a small ball with negative charge. The point P is outside the big ball, so we regard two balls (one positive and one negative) as point charge
and calculate EP = (k * (8 / 7) * q) / R ＾ 2 - (k * (1 / 7) * q) / (R-R / 2) ＾ 2
and prove that if the electric field in the cavity is the same, it is still regarded as two balls, only the point is in two balls. When processing, cancel any point t in the ball, and note that the vectors from the center of two balls to t are R1 and R2 respectively. The rest of the processing needs to draw, you do it yourself, if you have studied physics competition is not difficult.
and calculate EP = (k * (8 / 7) * q) / R ＾ 2 - (k * (1 / 7) * q) / (R-R / 2) ＾ 2
and prove that if the electric field in the cavity is the same, it is still regarded as two balls, only the point is in two balls. When processing, cancel any point t in the ball, and note that the vectors from the center of two balls to t are R1 and R2 respectively. The rest of the processing needs to draw, you do it yourself, if you have studied physics competition is not difficult.
7. 15MR²/32
8. If the same small circle is g out on the other side of symmetry, the center of gravity will still be at the center O of the big circle
the area of the excavated small circle is π( R/2)^2= π R ^ 2 / 4
the area of the big circle is π R^2-2 π R^2/4= π R ^ 2 / 2
area of small circle excavated: area of large circle of two small circles excavated= π R^2/4):( π R ^ 2 / 2) = 1:2
center of gravity O & # 39; The distance from the center of the big circle is R / 2 * 1 / (1 + 2) = R / 6
the area of the excavated small circle is π( R/2)^2= π R ^ 2 / 4
the area of the big circle is π R^2-2 π R^2/4= π R ^ 2 / 2
area of small circle excavated: area of large circle of two small circles excavated= π R^2/4):( π R ^ 2 / 2) = 1:2
center of gravity O & # 39; The distance from the center of the big circle is R / 2 * 1 / (1 + 2) = R / 6
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