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From the figure, we know that $F_{net} = mg\sin\theta$. Now, this force $\vec{F_{net}}$ is in the direction of the velocity $\vec{v}$ of the bob, both are tangent to the path. Therefore, the net acceleration $\vec{a_{net}}$ has no component perpendicular to the path, that is along the length $l$. I read that if acceleration is in the direction of velocity, then a body must be moving in a straight line, but such is not the case. Why? Also the bob is moving in a circular path and it should be experiencing centripetal force. What might be providing that force? The tension in the string is cancelled by the component of gravity parallel to the string.

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    $\begingroup$ Why would the tension in the string be cancelled by gravity???? Is gravity somehow pointing upwards? $\endgroup$ – CuriousOne Mar 24 '16 at 4:25
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    $\begingroup$ Tension is pointing along the string upwards, and the component of gravity perpendicular to the path points opposite the direction. $\endgroup$ – Omar Abdullah Mar 24 '16 at 5:09
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    $\begingroup$ Part of the tension is the reaction to the force of gravity parallel to the string, is that what you meant by "cancel"? I misunderstood that. In that sense we don't have to worry about that component of gravity as long as the string is under tension. The second part is the centripetal force that keeps the mass on the tangential motion. $\endgroup$ – CuriousOne Mar 24 '16 at 5:19
  • $\begingroup$ So $T$ is not cancelled, that is completely, by the parallel component of gravity (component along the string)? My book argues as follows: "since there is no motion along the string, the net force acting in the direction of the string is zero. This is possible if the component $mg\cos \theta$ balances the tension $T$." $\endgroup$ – Omar Abdullah Mar 24 '16 at 5:26
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    $\begingroup$ @Omer, your book sounds wrong to me. $T-mg\cos{\theta}$ acts towards O and must be greater than 0 to keep the pendulum from going off in a straight line. It causes centripetal acceleration. $T-mg\cos{\theta} = m(\dfrac{v^2}{l})$ $\endgroup$ – Ameet Sharma Mar 24 '16 at 5:39

The picture is valid, if nothing is moving. Otherwise, there has to be some centripetal force to the origin, since there obviously is acceleration on a curved path. The net force has a tangential component (if we are not in the deepest point) and a radial component (if we are not in the highest). This comes from an increased tension of the string.

Your book is just wrong: what matters is not, whether there is motion along the string, but whether there is acceleration -- which there is.

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    $\begingroup$ So the net force is not $mg\sin \theta$? It is then $F_{net} = \sqrt{(mg\sin \theta)^2 + (m\frac{v^2}{l})^2}$. $\endgroup$ – Omar Abdullah Mar 24 '16 at 6:14
  • $\begingroup$ @OmarAbdullah, that looks right to me. That's the net force acting on the bob. $\endgroup$ – Ameet Sharma Mar 24 '16 at 6:23
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    $\begingroup$ So, would it be correct to say that $T - mg\cos\theta$ is causing the acceleration that causes change in direction of velocity and $mg\sin\theta$ is causing acceleration/deceleration in the velocity that causes the pendulum to swing back and forth? $\endgroup$ – Peeyush Kushwaha Mar 24 '16 at 7:07
  • $\begingroup$ @Peeyush Kushwaha, yes, correct, you can split the net force in these two components. $\endgroup$ – Ilja Mar 24 '16 at 18:20

To understand the physics here we should first consider the assumptions made, and perhaps try to justify them with some physical arguments.

Assumptions of this simple pendulum:

  • Newtonian physics apply. (ie any non-classical effects are negligible)
  • The "string", or perhaps more accurately rod, has a fixed length $l$. (That is the string remains under tension but does not stretch)
  • The string is massless. (or is negligible next to the mass of the ball)
  • The system is frictionless, leaving only gravity (taken to be constant) and tension to act on the ball.

The string keeps the ball at a fixed distance $l$ from the pivot and hence the ball moves to trace out the arc of a circle. Knowing this we use Newtons Laws to resolve the forces involved. The acceleration due to gravity is taken to be $g$ hence the force (acting down) on the ball is $F_g = mg$, where $m$ is the mass of the ball. Decomposing this gravitational force into radial and tangential components we arrive at the expressions given in the diagram: $$F_{\text{radial}} = mg\cos \theta\ \text{ and }\ F_{\text{tangent}} = mg\sin \theta.$$ The tension $T$ in the string acts only radially and since it does not stretch, has magnitude equal to the sum of the centripetal force and the radial component of the gravitational force. The force necessary to keep an object in circular motion in free space (the centripetal force) is given by $$F_c = \dfrac{m v^2}{l},$$ where $v$ is the instantaneous velocity of the ball. Thus $$T = \dfrac{m v^2}{l} + mg\cos\theta.$$ So the tension in the string varies as the ball speeds up and slows down along it's trajectory. It is strongest when $\theta=0$ (at the bottom of the pendulum) since at this point the string is in direct opposition to gravity, and it is weakest when $v=0$ (at the top of the swing) as this is when the centripetal force vanishes and the radial component of gravity is at it's lowest.

It's worth thinking about the assumptions we've made and when they fail (for example is the string really always under tension?). When is this a useful model and how could we adapt it to account for the trickier edge cases?

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    $\begingroup$ Re: Your second assumption. A difference between a string and a rod comes apparent, if we give the bob enough initial velocity that it can climb up so that $\theta>90^\circ$. Assuming the bob stops there, then a string will have no tension, allowing the bob to drop straight down. But a rod would have "negative" tension, and push the bob to keep it on a circular path. Of course, with a high enough initial velocity even a string would remain taut all the way through the full circle. $\endgroup$ – Jyrki Lahtonen Mar 24 '16 at 13:26

Note that a (simple) pendulum performs (simple) harmonic motion and oscillates about it's mean position. For circular motion, the force must always be directed towards the center (centripetal). If you draw the force vectors at different positions (one position is shown in the image) as the bob moves note that the force is not always along the radial direction, i.e. towards O.

The image that you have posted likely depicts the extreme position of a pendulum (as $F_{net} = mg \sin\theta$), where the bob is momentarily/instantaneously at rest (another way to think here would be that the system has all potential energy at this very instant and no kinetic energy). At this very instant, the bob's velocity is zero and hence there is no centripetal force. However the tangential force (mg$\sin\theta$) will drive the bob out of that position and accelerate it to some non-zero speed. At this new position, the tension in the string/thread will now perform two functions-

  1. Balance the mg$cos\theta$ / perpendicular component of gravity.
  2. Provide the centripetal acceleration.

Hence the changing value of tension in the string cancels the component of gravity parallel to the string and provides the centripetal force.


What might be providing that force? The tension in the string is cancelled by the component of gravity parallel to the string.

I think a correction to the force diagram is required.

One should show the tension in the string and along the string a force called centripetal force is necessary to keep the bob in a circular path of radius equal to length of the pendulum.

No doubt the tangential net force is driving the pendulum.

So, T-mg Cos (theta) = centripetal force(provided for keeping the body on its

circular path ). The force of tension varies and its maximum at the lowest point ;

Suppose one releases the bob from horizontal position i.e. theta= 90 degrees then the tension at the lowest point comes to about 3mg .


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