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Using two different approaches, I appear to receive contradictory information about the tension force in a simple pendulum.

Under the idea of a centripetal force, the Tension - component of $mg$ in that direction(i.e. line of action of tension force) $=$ centripetal force to provide the 'circular motion' for small $\theta$ in a simple pendulum

However, resolving forces in that direction and using $F= ma$, since the string length is fixed, the bob at the end of the string doesn't change 'height' in direction of the tension force so surely

Tension $=$ Component of $mg$ in that direction

Any help would be appreciated!

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  • $\begingroup$ Your second approach is bit unclear to me . "However, resolving forces in that direction and using F= ma," , which forces r u resolving and what is that direction? $\endgroup$
    – Bhavay
    Commented Aug 13, 2020 at 21:16
  • $\begingroup$ The pendulum is never in static equilibrium, so you can't "resolve forces" and ignore the mass x acceleration term in the equation of motion. $\endgroup$
    – alephzero
    Commented Aug 13, 2020 at 22:35
  • $\begingroup$ $F=Mgcos\theta$ now if $\theta$ is small then you are doing the same thing it does have the acceleration. $\endgroup$ Commented Aug 14, 2020 at 3:21

2 Answers 2

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You can simply imagine a circular motion. In a circular motion the distance of a particle describing the circle from the centre remains constant. This distance is the string of the pendulum (which doesn't slack) but still you have some acceleration towards the centre. What happens is that the particle constantly tries to fall towards centre but is taken back due to a velocity tangentially at any given point. So instead of simply falling to the centre it rotates around.

Similarly, in the case of a pendulum though the length remains same. It still has some net force towards the centre and Bob does try to fall towards the pivoting point but instead of that, it undergoes the pendulum motion due to the presence of a tangential component of $mg$ (perpendicular to the component which is along the direction of tension) which helps the Bob to describe a kind of circular shape.

So stick with the first one , i.e. : $T$ - component of $mg$ in that direction= centripetal force and

$F_{net}$ is not equal to zero (along $T$) $\implies T$ is not equal to the component of $mg$ along this direction

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    $\begingroup$ Okay that makes sense, but is there a reason why the forces can't be resolved in that situation- is it as a result of the rotational motion. Is there conditions to resolve the forces like I did. Thanks $\endgroup$ Commented Aug 14, 2020 at 8:59
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    $\begingroup$ You are close now.....imagine suppose you are holding the pivoting point now without rotating the pendulum,i.e, the pendulum is just hanging down, in that case your thought that T= mg is correct. We are in fact resolving the force mg even when in rotation but the thing is that both T and the 'resolved' component of mg along T work together to produce a net 'centre-seeking' force. Hope it clears your concept! $\endgroup$ Commented Aug 14, 2020 at 11:46
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Just because there is no change in length of the string doesn't mean there is no acceleration in the radial direction. If you understand circular motion properly, you'll see that the particle undergoing circular motion is always accelerated towards the centre, even if the radius remains constant. The acceleration is such that the direction of velocity changes, as a tangent to the circular path. Thus when you apply $F=ma$, you have to consider not only the tensile and gravitational forces, but also the centripetal acceleration.

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  • $\begingroup$ Ok thanks that makes generally sense. Acceleration is change in velocity over time. This centripetal acceleration is due to the linear speed of the particle changing direction right? $\endgroup$ Commented Aug 14, 2020 at 8:58
  • $\begingroup$ Centripetal acceleration does depend on v(linear tangential speed) as v² but it doesn't arise as a result of the "linear speed". It is just due to the presence of a 'centre seeking force' towards the pivoting point. In your pendulum case, the acceleration which is actually due to the change in linear speed(rather say which changes the linear speed) is the mgsin@ component (component perpendicular to the tension) $\endgroup$ Commented Aug 14, 2020 at 11:33

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