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Suppose a particle with mass $m$ is whirled at instantaneous speed $v$ on the end of a string of length $R$ in a vertical circle. Let $\theta$ be the angle the string makes with the horizontal. I know that at all points, the equation: $$T = \frac{mv^2}{R} - W \sin{\theta}$$ must be satisfied. What I'm not so sure of is the condition required for cicular motion in itself to take place. I first thought it's $\frac{mv^2}{R} \geq W \sin{\theta}$, since negative tension makes no sense. But then I realized that zero tension is equivalent to the absence of a string attached to the mass (i.e. no circular motion). And then I read that tension can be zero at the top, and the mass would still be in circular motion. What's going on here? Is tension allowed to be zero only at the top? I'm confused. Also, this made me think: what if I were to cut the string at any point, what would happen then? The answer is pretty obvious: the trajectory of the particle would be parabolic. The answer wasn't that obvious when I started to think about the situation in terms of polar coordinates though. The component of the weight of the particle in the radial direction $(W \sin{\theta}$ always points towards the origin (at least in the upper half of the circle). It is always radial. Why doesn't circular motion occur? In other words, why is $W \sin{\theta}$ not a good candidate, unlike tension? Please try to keep all arguments in terms of forces (not energy).

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  • $\begingroup$ Still no answers? $\endgroup$ – 0MW Mar 13 '15 at 22:14
  • $\begingroup$ There are 3 cases here.firstly if T<0 then the motion becomes simple harmonic .if T becomes 0 in between then the particle will follow its subsequent motion as a projectile and finally if T>0 then it is motion in a vertical circle.By using FBD you may realise that T<0 is possible only if the particle is at 0<theta<90 . $\endgroup$ – shrey Jun 1 '15 at 8:39

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