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First of all, if I'm not mistaken, there's a small error in your title. With the earth as a reference, the tension should be m 2*r. Don't make a mistake about the basic formula.

And then if you want to use one ball as a reference, two balls are not actually relatively static!

Your understanding should be that there are observers standing on one ball and facing the other. In his opinion, the other goal has not changed in relation to his position. So you say that the two balls are relatively stationary.

But in the process, you're not choosing a ball frame of reference! Rather, it is the "observer" frame of reference. The observer's angle is always spinning with the ball!

In general, the reference frame we choose should be translational only.

So it needs to be changed. For example, the observer is still standing on the ball, but the difference is that he is always facing north relative to the earth! At this point, he will find that the other ball is also spinning around him.

What you are asking for is t. Since the frame of reference itself also has a centripetal acceleration a. So we need to add the inertial force ma to the two balls (opposite to the direction of a) and for this ball chosen as the frame of reference, it is stationary, so there is t=ma

For the other ball, it does a circular motion with a radius of 2r, centripetal force = m 2*2r = ma + t = 2t

Solve t=m2*r

This is obviously more complicated, but it can certainly be correct.

It's a bit much, if you don't know what you say, you can ask.

When selecting a reference frame for a force analysis, it is important to specify whether the selected reference frame is an inertial or non-inertial reference frame.

When a sphere is selected as a reference frame, it moves at variable speed relative to the inertial frame, that is, in this non-inertial frame of reference it is equivalent to an equivalent force due to the conversion of the reference frame, this force is called the inertial force, its magnitude f=ma, and the direction is opposite to the direction of acceleration.

First of all, the tension is m 2r, we know that the two balls are connected by the rod to make a uniform speed translation rod is not to exert force on the ball, the two balls are also relatively stationary, but if it is rotated around the center of the rod is powerful, relatively stationary and uneven speed is also powerful, why, because the circular motion has centripetal acceleration, and the variable speed motion also has acceleration, so it can only be judged according to its trajectory and state.

The net force of the system: f=(m+m)a

From the impulse theorem, from the beginning to the end: f*(t1+t2)=mv+m*0-(m*0+m*0).

Solution: v=(m+m)(t1+t2)a m

The velocity of t1 is v1=at1 at all times. The time t2 is known by the momentum theorem, (m+m)v1=m x 0+mv2. So v2=(m+m)at1 m.

Newton's Second Law and Kinematic Formulas.

Careful analysis is just a bit troublesome and not difficult.

The answer is c. Raindrops fall from a high altitude from a standstill, so the starting point of the image must be at the origin point (excluding a and d).

As the raindrops fall faster and faster, the resistance also increases, and when the resistance is the same as the gravitational force of the raindrop (opposite direction), the two forces are balanced, and the raindrops move at a uniform speed (the right side of the image of the answer c is close to a straight line. ）

c g does not change, and the drag increases as v increases, so the acceleration becomes smaller.

Choose C, the speed of the raindrop will definitely increase when it is constant, and then it will stabilize when the drag is equal to gravity.

The acceleration is reduced to zero all the way through.

7.If the resultant force is fixed at one end of the rod, then the resultant force is 1n, and the torque generated at each point of the rod perpendicular to the rod is farther away from the fixed point, the greater it is.

8.A couple, a pair of two forces of equal magnitude and opposite directions, but whose lines of action are not in the same straight line, constitute a couple. Therefore, the sum of the projections of the couples on any coordinate axis is zero.

9.In motion on a plane, the translational degrees of freedom are 2 and the rotational degrees of freedom are 1, so 12 independent equilibrium equations can be established.