# Centripetal force of a rotating rigid body?

Consider someone pushing a roundabout in a playground. Initially the roundabout is stationary, but when it is pushed, it rotates with increasing rotational speed.

The force of the push is balanced by the reaction force exerted by the support at the centre of the roundabout. The forces are equal in magnitude and opposite in direction, so the roundabout is in translational equilibrium. But they have different lines of action, so there is a resultant torque, causing the playground to rotate and have angular momentum.

Okay, my question is, how about the centripetal force that exists whenever there is circular motion? Where does/would it come from?

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– Qmechanic♦ Apr 25 '12 at 9:50
The 2 threads linked above are discussing how centripetal force is a resultant force and not a real force in the sense that tension, weight, etc. are. My above question on the other hand is about the difference between a particle in revolution versus a body in rotation, or how exactly to relate the centripetal force in the former to the context of the latter. So the 2 linked threads are not directly related to this discussion. Just to clarify. :-) – Ryan Apr 25 '12 at 16:22
– Ryan Apr 25 '12 at 17:58

## 3 Answers

Consider the simpler system of a mass in 2D, connected to the origin by a massless rod that is free to rotate about the origin. When you exert a force perpendicular to the rod on the mass, the mass exhibits circular motion. In this case, the centripetal force needed comes from the tension exerted on the mass by the rod.

A similar situation happens for the roundabout: tensions in different parts of the roundabout act on each other to give the necessary centripetal force.

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Thanks! Since the roundabout is rotating, there should be a net force towards the rotational axis, right? I know I'm missing something fundamental... – Ryan Apr 25 '12 at 10:41

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When you sum the forces up, the result describes what the center of mass is doing, not what the individual parts of the roundabout are doing. In this case, the net force on the roundabout is zero, since its center of mass is stationary. To see the circular motion, we find the net force on individual parts of the roundabout. – leongz Apr 25 '12 at 10:53

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Thank you, Leongz. I've finally wrapped my mind around it. So in summary, the net force towards the rotational axis exists for each "particle" of the rounadbout, but not for the roundabout as a whole body. Cheers. – Ryan Apr 25 '12 at 16:29
@Ryan Exactly. This is the general idea. – Pygmalion Apr 25 '12 at 16:30
Stackexchange is the best invention since water. – Ryan Apr 25 '12 at 16:35

In order to have a centripetal force, you must have mass that rotates around certain point. You should be more specific with your question, that is, you must tell us which mass is rotating and then we can tell you which centripetal force is responsible for that rotation.

Here is a more complete explanation on where does the centripetal force come from: Let's suppose we are standing in intertial frame of reference. As first Newton law states: if no force is exerted to a body, velocity of the body remains constant. Now what about rotating? In rotating velocity is not constant!

OK, velocity is vector, and it is possible that the body rotates in a way that the magnitude of the vector is constant. However, if the body is rotating, the direction of the vector of the velocity is changing! Therefore, some force must exert on the body, must force the body to rotate. It turns out that the force, that changes the direction of the vector of the velocity is directed toward the center of the rotation, and therefore we call that force centripetal force (petere in Latin means: to make for, tend to get to), i.e. force that tends toward center.

For specific explanation provide specific case.

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Oh okay, that makes perfect sense. But yet, when considering the body as a whole (instead of an individual particular part of the body), shouldn't there be a net force towards the axis of rotational motion? – Ryan Apr 25 '12 at 10:39

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No. Take a look at two equal parts of the roundabout on the oposite sides of the axis. There are two centripetal forces of the equal magnitudes. But since they are both pointing toward the center, they have oposite directions and they cancel out when you look roundabout as a whole body. – Pygmalion Apr 25 '12 at 11:23
If roundabout is not symmetrical and you have larger mass on one side of the axis than on the other side, one of two centripetal forces will be greater than the other and then there will be net force acting on the axis. Axis of the roundabout will wobble. – Pygmalion Apr 25 '12 at 11:29
Thanks, Pyg, got it. :) – Ryan Apr 25 '12 at 16:06

The centripetal force isn't a "new" force that comes out of nowhere. It's made up of normal forces.

Allow me to explain:

For any sort of acceleration, via $\vec F=m\vec a$, you need a force, right? So basically, if you were accelerating a rock with some rope, the corresponding force is the rope tension. If a ball is falling on Earth, the corresponding force is gravity. You can even try pulling a ball down on Earth, and you get a mixture of forces.

Now, note that velocity is a vector. In UCM (uniform circular motion), you have a body with constant speed, but the direction of its speed varies.

In the above diagram, the green arrow is the initial velocity vector. The blue arrow is the velocity after a split second. The red arrow is the acceleration vector required to change the velocity thus.

Now, by $\vec F=m\vec a$, we need a force to make this happen. So, which force is it? It depends.

If you are whirling the stone in a horizontal plane via a rope, this force is the tension force. If you are whirling it in a vertical plane, it is the combination of gravity and rope tension. It the ball is rolling in a bowl, it is the combination of gravity and reaction(normal) force. If this is a planet-satellite system, the force is just gravity.

A general term for this force is "centripetal force". CPF is a mixture of pre-existing forces, depending on the situation, as explained above. Using some calculus, we can prove that $\mathrm{CPF}=\frac{mv^2}{R}$, where $m$ is the mass of the particle, $v$ is the instantaneous velocity, and $R$ is the radius of curvature (just the radius in case of circular motion).

So it doesn't "come from" anywhere--it's not a new force. It's just a name for pre-existing forces when they create circular motion. That's all.

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I mean, shouldn't there be some sort of net force in the centripetal direction? In my roundabout example, aren't both forces tangential (in opposite directions) to the circular motion? – Ryan Apr 25 '12 at 10:35

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@Ryan. There is. The tangential pushing accelerates the roundabout and puts it in UCM in the first place. Then, the metal of the roundabout does the same thing as a rope-it keept is in UCM. – Manishearth♦ Apr 25 '12 at 10:40

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@Ryan: There is a net CPF on every individual small element of the roundabout. Overall, the CPFs add up to zero. CPF only applies to small elements-- for a larger body the internal forces come into the picture (these can be calculated easily, though). Also note: whenever a constant-magnitude force is centripetal, i.e. it changes its direction to point at a particular point, then there will be no net translational motion. – Manishearth♦ Apr 25 '12 at 10:50
With reference to the yellow paragraph in my question, there are two obvious forces (neither of which point toward or away from the rotational axis), and their sum is also the final resultant force on the body. So the metal of the roundabout is like an imaginary string keeping each particle on the circumference of the body in circular motion. So why isn't there a resultant force pointing towards the rotational axis? I'm so confused. – Ryan Apr 25 '12 at 10:56
Okay, let me try to digest your explanation above that the CPF's add up to zero. Need to be away from the computer for a few hours. Thanks Manis! – Ryan Apr 25 '12 at 11:00
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