# Why does the centripetal force not given in the Equation (I)?

Example 4.1(An Introduction to Mechanics, 2E,Kleppner and Kolenkow)

The Bola is used by gauchos for entangling their cattle. It consists of three balls of stone or iron connected by thongs. The gaucho whirls the bola in the air and hurls it at the animal. What can we say about its motion? Why didn't the author consider centripetal force on each objects?

Equation (I) should be

$$\frac{d\vec{P}}{dt}=m_1g+m_2g+m_3g+\frac{m_1 v_1^2}{r}+\frac{m_2 v_2^2}{r} +\frac{m_3 v_3^2}{r}$$ instead of the given one? Please help me to understand the problem.

• The centripetal forces are the tensions along the thongs, so they all are internal forces, i. e. forces from parts of the system to other parts. Internal forces do not change the system momentum (this is very likely proven earlier in the chapter the example is). – Rafa Budría Aug 26 '20 at 16:27

## 1 Answer

Because those are virtual forces, if you write them in a rotating reference frame. From a fixed reference frame, or one moving with a constant velocity, you need to replace those forces with $$m\vec a$$ instead. Another way of thinking is asking yourself what (or who) is generating the centripetal force? If I look at an object moving in a circle I can write $$ma=m\frac{v^2}r=F$$ Here $$F$$ might be a tension in a string or gravity for the planets. If I am in a rotating reference frame where I see the object fixed, even if I have a force $$F$$ acting on it, I conclude that there must be something else there as well to cancel $$F$$. But that force is not real, since it only exists in the rotating reference frame.

• You mean virtual forces wouldn't count! But when we consider a car moving in a banked road or earth orbitting around the sun. In these model we take the radial component of forces equals to centripetal force. I am confused. Can you make your answer clearer? Thank you in advance. – Unknown x Aug 26 '20 at 2:16
• Let's consider the Earht moving around the Sun. If you watch from a fixed reference frame, there is only gravity. Then you write $$m\vec a=-\frac{GmM}{r^2}\hat r$$As you can see, there is no centripetal force. If the trajectory is a circle, then $a=v^2/r$, and that's it. Now if you consider in a rotating reference frame such that one axis is always pointing along $\hat r$, you have $a=0$. So you need to introduce a fictional force, the centripetal one, so it will cancel the gravity. But this force exists only in this rotating reference frame. – Andrei Aug 26 '20 at 2:39
• I comsidered earth moving around a sun from earth. I agree I can see only gravitational force now. In the second case, I am coming out from the earth and stand some where in the space, In that case, I can see earth revolve around a sun(assume it is revolve in a perfect circular motion). Here due to uniform motion there will be an acceleration towards the centre. But in the problem, I asked from where should I observe? I am observing the bola. Right? Here bola is rotating. So, shouldn't I consider the centripetal force? – Unknown x Aug 26 '20 at 3:02
• If you are fixed and see things rotating, then you see acceleration, so no centripetal force. If you are rotating and see things "fixed", then you have centripetal force – Andrei Aug 26 '20 at 3:19