A train is rounding a curve, that is part of a circle with a radius R. A pendulum on the train swings out to an angle θ throughout the turn. What is the speed of the train?

I'm not sure what the centripetal force on the pendulum is here. There is a tension force pulling it up and a gravitational force pulling it down. What other forces are there? It's not touching anything so it can't be a normal force or a static friction force. Once I know what this force is, I can easily do the rest of the problem, but I have no clue what it is.

  • $\begingroup$ "I'm not what the centripetal force on the pendulum is here" - sorry, what? Perhaps an edit? The force in question is clearly centripetal, from the rails. $\endgroup$ – Maury Markowitz Oct 4 '16 at 19:16
  • $\begingroup$ @MauryMarkowitz Sorry, I missed a word. "I'm not sure what the centripetal force on the pendulum is here" is what I meant to say. And I'm asking what the force is on the pendulum. The rails can't be exerting a normal force on the pendulum because the normal force is a contact force. The only forces I know of acting on the pendulum are an upward tension force and a downward gravitational force. So where is the centripetal force coming from? $\endgroup$ – RothX Oct 4 '16 at 19:28
  • $\begingroup$ Ahhhh. Ok, but the rails are contacting the train. The train is contacting you. You are contacting the pendulum. Is that what you mean? This is assuming you're holding it between thumb and finger, but it works just the same if you tie it to the roof. So then, the force is ultimately electromagnetic, but I don't think that's what you mean. $\endgroup$ – Maury Markowitz Oct 4 '16 at 19:36
  • $\begingroup$ Well sure, it's ultimately electromagnetic, but yeah, that's not what I mean. And also, I'm not on the train, it's just hanging there. Is it true that the only forces on the pendulum are gravity and a tension force? If so, where is the centripetal acceleration coming from? $\endgroup$ – RothX Oct 4 '16 at 19:43
  • $\begingroup$ If "its just hanging there" means from the roof of the train, then that's where it's coming from. The rails press on the train, the train presses on the top of pendulum rope. $\endgroup$ – Maury Markowitz Oct 4 '16 at 19:57

Forget about the rails. Absolutely irrelevant. The only forces on the pendulum bob are tension and gravity. Tension is a contact force, gravity not. No other forces on the bob.

The centripetal force is not a separate force. This is a source of great misunderstanding. Better to think in terms of Newton's second law: $$a=F_\mathrm{net}/m$$ In the case at hand, $F_\mathrm{net}$ is the sum of the force of gravity and the tension force. $a$ is the acceleration that is observed, centripetal acceleration in this case. You have the force on the right, and the resulting acceleration on the left.

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  • $\begingroup$ I thought that they might be the only two forces, but how does the tension force have a horizontal component? And how can I find it? $\endgroup$ – RothX Oct 4 '16 at 20:18
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    $\begingroup$ You told us it has swung out at an angle ... if you can't figure this out you should review the decomposition of vectors. $\endgroup$ – garyp Oct 4 '16 at 20:45
  • $\begingroup$ Well, yeah, I know how that works. I guess my question isn't clear. What is caused it to swing out at an angle? And why is it staying there? Because the x-component of the tension force should be pulling it back, right? This is why I felt there might be another force. $\endgroup$ – RothX Oct 4 '16 at 20:52
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    $\begingroup$ You are trying to view the system from a frame of reference attached to the pendulum. That frame of reference is not an inertial frame (it is accelerating) so Newton's law's are not valid. You have two choices: Consider the system from the point of view of a stationary observer on the ground and use Newton's second law (recommended) or consider the system from a frame attached to the pendulum and add a fictitious force (not recommended until you understand the recommended method). For the recommended method write it all out, and don't forget to include the acceleration. $\endgroup$ – garyp Oct 4 '16 at 21:16
  • $\begingroup$ But how can I figure it out without the mass? v=sqrt(ar) F=ma I know r, so 2 equations and 4 variables? Where can I use the angle? $\endgroup$ – RothX Oct 4 '16 at 23:21

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