# Tension, Centripetal force and Simple-Pendulum

### At position 3

We've to find tension in the string (ideal).

The answer is: $$T = mg\cos\theta$$

But my doubt is, how can Tension at 3 balance the radial component of gravity.

If that were so the centripetal force should be zero but that's certainly not the case since the mass is exhibiting circular motion.

• Isn't position 3 supposed to be the extreme right hand limit of the bob's swing? In that case the bob is instantaneously at rest, so there is no centripetal acceleration. Commented Jul 26, 2023 at 15:12
• @PhilipWood , so you're saying $$F_C =\frac{ m{v_{\text{inst}}}^2}{r}$$ ? Is this true even if there exists a tangential acceleration in the circular motion? Commented Jul 26, 2023 at 15:16
• Yes, that's what I implied. And yes, even if there is a tangential acceleration (which there is). Commented Jul 26, 2023 at 15:17
• @PhilipWood , Is this true even if there exists a tangential acceleration in the circular motion? Commented Jul 26, 2023 at 15:18
• Yes, even if there is a tangential acceleration – which there is at position 3. Commented Jul 27, 2023 at 6:41

In general, Newton's second law in polar coordinates is given by

$$\mathbf F=m\left(\ddot r-r\dot\theta^2\right)\hat r+m\left(r\ddot\theta+2\dot r\dot\theta\right)\hat\theta$$

In the case of the pendulum, $$\dot r=0$$ and $$\ddot r=0$$, and so we are left with

$$\mathbf F=-mr\dot\theta^2\hat r+mr\ddot\theta\hat\theta$$

Also for our pendulum, we have $$\mathbf F=\left(mg\cos\theta-T\right)\hat r-mg\sin\theta\,\hat\theta$$

Therefore, we can determine from Newton's second law that

$$mg\cos\theta-T=-mr\dot\theta^2$$ $$-mg\sin\theta=mr\ddot\theta$$

At a point in time where $$\dot\theta=0$$, we must then have $$T=mg\cos\theta$$ at that point in time as well.

But my doubt is, how can Tension at 3 balance the radial component of gravity.

If that were so the centripetal force should be zero but that's certainly not the case since the mass is exhibiting circular motion.

We have circular motion, but not uniform circular motion. The centripetal acceleration varies during the motion of the pendulum, and as we can see is equal to $$0$$ at the ends of the swings.

But my doubt is, how can Tension at 3 balance the radial component of gravity. If that were so the centripetal force should be zero but that's certainly not the case since the mass is exhibiting circular motion.

The tension force balances the radial component of gravity making the centripetal force zero at the extreme horizontal position of the bob, not at position 3.

In general, the centripetal force is given by

$$F_{C}=\frac{mv^2}{R}$$

Where $$v$$ is the speed of an object of mass $$m$$ in circular motion with radius and $$R$$. For uniform circular motion the speed of the object is constant the centripetal force is constant. That is not the case for the pendulum where the speed of the bob and the centripetal force varies with the angular displacement. For the pendulum $$v$$ is the instantaneous speed.

For the pendulum, where $$R$$ is the length of the string, the centripetal force is the net radial force acting on the bob, or

$$\frac{mv^2}{R}=T-mg\cos\theta$$

At position 2 the speed of the bob is a maximum of $$v=\sqrt {2gh}$$ where $$h$$ is the maximum vertical height of the bob above point 2 Ar this point the centripetal force is a maximum.

Hope this helps.

• Shouldn't this be the expression: $$F_C = T - mg\cos\theta$$ ? Commented Jul 26, 2023 at 15:11