Suppose I have a simple pendulum and I want to calculate its acceleration when the bob reaches the maximal angle. I usually choose my axes such that the y-axis will be parallel to the rope. Then the acceleration would be $g \sin\alpha$ (Because: $mg\sin\alpha=ma$). However, if I choose the y-axis to be parallel to the gravitational force ($mg$), the acceleration equals to $g\tan\alpha$ (Because: (1): $T\cos\alpha=mg$ and (2): $T\sin\alpha=ma$ where $T$ is the tension force (centripetal force) and $\alpha$ is the given angle). Obviously, these two accelerations aren't the same. So my question - which one is right? And why the other one is wrong?


The total gravitational force is $mg$ and this force is the hypotenuse of a triangle whose sides are $mg\cos\alpha$ in the direction of the rope and $mg\sin\alpha$ in the orthogonal direction. So the latter, $mg\sin\alpha$, is the force (mass times acceleration) along the circular orbit because the centripetal part of the force is cancelled by the string's tension.

My guess is that instead of dividing the gravitational force into the two orthogonal components, you are trying to divide the tension force into two components and identify one of these two components with the gravitational force.

The plan wants to divide the tension $T$ into two directions – I suppose it's the horizontal and vertical direction. In the vertical direction, you think to have the cancellation between $mg$ and $T\cos\alpha$ and in the horizontal direction, between $T\sin\alpha$ and $ma$.

However, you forgot that in the vertical direction, there's a contribution to the force from the acceleration, too. So the right conditions in this coordinate system are actually $$T\cos\alpha = mg-ma\sin\alpha, \quad T\sin\alpha = ma\cos\alpha$$ Note that instead of $ma$ from your formulae, I had to write $ma\sin\alpha$ and $ma\cos\alpha$ because the acceleration is a vector in a direction that is neither vertical nor horizontal. Multiply the first (correct) equation by $\sin\alpha$, the second one by $\sin\alpha$, and subtract them in order to eliminate $T$. You will get $$ 0 = T(\sin\alpha\cos\alpha-\cos\alpha\sin\alpha) =\\ = mg\sin\alpha - ma\sin^2\alpha-ma\cos^2\alpha=mg\sin\alpha-ma $$ which gives you $a=g\sin\alpha$ again. In other words, you have forgotten to realize that the acceleration $a$ is in a direction tilted by $\alpha$ so if you rewrite the vectors in the horizontal and vertical components, you have to rotate it into $a\sin\alpha$ and $a\cos\alpha$ with the right signs.

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    $\begingroup$ Thank you. I indeed forgot about the acceleration vector. But you could skip the mean part about me misunderstanding "something important about physics" or a whole paragraph about how I'm wrong, by simply clearing the things out. You know, demotivating isn't a contribution. Thanks anyway. $\endgroup$ – hsjk04 Mar 26 '13 at 17:32
  • $\begingroup$ I edited that bit out of the answer. Just be careful with that in the future, Lubos. $\endgroup$ – David Z Mar 26 '13 at 17:37
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    $\begingroup$ Sorry, I just wrote that it isn't possible to properly understand a wrong derivation without making errors which is important for outlining the proper status of the rest of my answer - it is about empathy, not about physics because physics may only objectively talk about valid arguments, not about invalid arguments that may look valid because of people's errors. Could you please return the formulation in the answer, David, and avoid similar degradations of users' answers in the future? Thanks. $\endgroup$ – Luboš Motl Mar 26 '13 at 17:42
  • $\begingroup$ Dear Lumo, you can roll back your answer yourself if I am not mistaken. What @DavidZaslavsky and Alec S have done, namely messing around with your answer was wrong, unneded and impertinent. And David should by no means have threatend you since there was absolutely nothing wrong with the original version of this answer, it was neither insulting nor something else ... ! $\endgroup$ – Dilaton Mar 27 '13 at 21:30
  • $\begingroup$ @Lubos it's a matter of civility, which as you know is an expected value from all contributors here. If I removed some essential physics content from your answer, go right ahead and edit it back in, as long as it stays free of personal attacks or insults or anything along those lines. $\endgroup$ – David Z Mar 27 '13 at 22:12

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