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Suppose an object is rotating in a circular motion, and we change increase its radius pulling it outwards in such a way that the net torque is always zero. In such a case, the angular momentum will be conserved so the velocity and the omega of the object will decrease. I wanted to know the exact mechanical reason behind this thing, that which force exactly slows the velocity of the object while it moves from smaller radius to the bigger one. And I got the answer from a video on YouTube: https://youtu.be/_WHRWLnVm_M (just skip to 12:00 to get what I am talking about) The part of the video basically says that the centripetal force starts decelerating the object when it jumps from one orbit to another (following a curve path), which causes the velocity to slow down.

The black line represents the path of the object, which isn't straight from one orbit to another but it follows a curve path. During the curve path, the blue line which is the centripetal force decelerates the object's velocity.

Now, I have 2 questions. One is that what exactly causes the centripetal force to change its magnitude here?. And other one is that, this only applies when the direction of the object is just changed in such a way that the it reaches the required orbit. What if I pull it with a radially outward external force, such that there is no net torque. That force must be more than the centripetal force which cancels this explanation for me. In that scenario how is the velocity decreased and the angular momentum conserved?

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Updated answer with correction about conservation of angular momentum. Also see the other question/answers referenced.

In an earlier answer I said for a particle in initially uniform circular motion, for a rapid change in the centripetal force angular momentum is not conserved; that is incorrect if the force is purely centripetal since it is always parallel to the moment arm to the particle. However, during the movement, the force is no longer normal to the velocity and as such does work on the particle. See Centripetal force: if radius decreases, does ANGULAR or TANGENTIAL velocity change?

The motion is most easily evaluated using polar coordinates. The accelerations in the radial and angular directions are $F_c = m(\ddot r - r \dot \theta^2)$ and $0= m(r \ddot \theta +2 \dot r \dot \theta)$ where $F_c $ is the centripetal force and $m$ is the mass of the particle. Angular momentum $mr^2 \dot \theta$ is constant.

A related problem where the centripetal force can change, is that of a body in motion subject due to the gravitational force from another body, if the motion is not circular, say an ellipse. See a physics mechanics textbook for an evaluation of this problem, such as Mechanics by Symon.

You asked about another force applied radially outward. The centripetal force is the net force in the radially inward direction, so your applied force changes the magnitude of the centripetal force.

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  • $\begingroup$ And what exactly causes the centripetal force to change its magnitude (decrease in this case) here? I don't see any force opposing it, if we change it slowly. $\endgroup$
    – Ani
    Commented Jan 15, 2023 at 22:46
  • $\begingroup$ For example, a particle on a string, letting the string out decreases tension, tension is the centripetal force. Pulling the string in increases tension (centripetal force). Or a body in an elliptical orbit around a fixed body subject to a changing magnitude of the gravitational force. In these cases, there is not force other than tension or gravity, respectively. $\endgroup$
    – John Darby
    Commented Jan 16, 2023 at 16:00
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It follows from the definition of angular momentum. I suppose that an sudden impulse was done on the object with mass $m$. After that, its velocity changes according to the net (centripetal) force on it: $$\mathbf L = \mathbf r \times \mathbf p = m\mathbf r \times \mathbf v$$

Making the derivative with respect to time: $$\frac {d\mathbf L}{dt} = m\frac{d\mathbf r}{dt} \times \mathbf v + \mathbf r \times \frac{md\mathbf v}{dt} = m \mathbf v \times \mathbf v + \mathbf r \times \mathbf F$$

Both cross products are zero because the vectors have the same direction.

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