Suppose we have a person sitting on a rotating chair,he is rotating with angular velocity w(omega) .Then he extends his arms outside,this changes the angular velocity to w'. My question is which force causes such a change? Which force causes the boy to change its angular velocity?


Ever stuck something on the blades of a fan to watch it get thrown around the room? Then you know that the rotating arms can accelerate a mass as it moves from the center to the edge.

If the mass is being given kinetic energy, then it must be coming from something and that is the rotating arms. So there is a force pair between the arms and the object. This force is accelerating the mass and decelerating the arms. (The motor on your room fan is probably powerful enough that you don't notice it in this case).

So the answer is that as the person's arms/hands are flung outward, they push backward on the body. This push slows down the rotation.

Here's another way to think of it. When the masses are fixed on the arm, the force is purely radial. The arms apply a centripetal force to keep the masses in uniform circular motion. This force is through the axis, so there is no change in rotational speed.

Immediately before release, the object and the arm share rotational velocity $\omega$, and both the object and the attachment point on the arm share a tangential linear speed $v$. After the object is released, it moves along the arm to a distance farther from the center. But it still only has linear speed $v$. At this new distance, that speed would give it a rotational velocity less than $\omega$. Because the arms and the object constrain each other, this means the arms must supply a force to increase the object's linear velocity, which is going to reduce the arm's linear velocity.

  • $\begingroup$ But if the arms are extended slowly. Then there wouldn't be any backward force. Right? Also the backward force would pass through the axis of rotation so there shouldn't be any torque right? $\endgroup$ – Lakshya Gupta Apr 24 '17 at 5:35
  • $\begingroup$ @LakshyaGupta Doing something slowly doesn't mean no force; it means less force but over a longer duration. $\endgroup$ – Steeven Apr 24 '17 at 8:23
  • $\begingroup$ Agreed but it would still pass through the axis of rotation. $\endgroup$ – Lakshya Gupta Apr 24 '17 at 8:32
  • $\begingroup$ The arms are not pushing the mass directly out, they are pushing them forward. So the force does not pass through the axis. In the ideal case, the force is normal to the radial. $\endgroup$ – BowlOfRed Apr 24 '17 at 14:59

It is perhaps easier to start with a simpler system?

Imagine that an unextended spring is tethered at the centre of rotation and at the other end of the spring there is a mass which is held in position on the rotating platform.
The mass is in a frictionless groove which is orientated radially outwards from the centre of rotation.

The tether is removed.

As the mass moves outwards the wall of the groove attached to the rotating platform exerts a tangential force on the mass, resulting in a torque about the centre of rotation, to increase the rotational speed of the mass.
At the same time the mass exerts a tangential force on the groove and the rotating platform (Newton's third law), resulting in a torque about the centre of rotation, which reduces the rotational speed of the groove and the platform.

Suppose the spring is its natural length (unextended) and the mass is tethered to the groove.
To keep the mass rotating it must be subjected to an inward force provided by whatever is holding the mass in position.
If the mass is now allowed to move freely it will move outwards and the spring will exert an inward force on the mass.
So the force on the mass due to the spring is inwards and the mass is moving outwards so negative work is done on the mass and the kinetic energy of the mass and the platform is reduced.

The spring has an outward force exerted on it due to the mass (Newton's third law) and the displacement is also outwards so positive work is done on the spring.
That positive work manifests itself as elastic potential energy stored in the spring.

For your more complex situation there are equivalent forces and displacements.


Let me change the person in the chair, for a mass attached to a string. Take a look to escalar definition of angular momentum $L=r\, p = r \, mv$ where $r$ is the string longitude and $p$ is the mass linear momentum. The angular momentum is a conserved cuantity. Then if, the string longitude changes, the linear momentum have to changes too for maintain the angular momentum as at the first. ($\omega = \frac{v}{r}$) The force that you are looking force is exerted by the boy, because he is changing the movement radius thus the centripetal acceleration.


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