- #1
PFuser1232
- 479
- 20
Suppose a mass ##m## is whirled at instantaneous speed ##v## on the end of an inextensible string of length ##R## in a vertical circle. Let ##\theta## be the angle the string makes with the horizontal, and let ##W## be the weight of the particle.
According to the equation:
$$T = \frac{mv^2}{R} - W \sin{\theta}$$
##\frac{mv^2}{R}## must be greater than or equal to ##W \sin{\theta}## since tension can never be negative.
But if both terms are equal, then the tension of the string is zero. The particle no longer follows a circular path if ##T = 0##, since this is equivalent to the absence of a string holding on to the particle, right?
So the condition for pure circular motion (no coriolis or linear radial acceleration terms) should really be:
$$\frac{mv^2}{R} > W \sin{\theta}$$
without an equal sign, is this correct?
According to the equation:
$$T = \frac{mv^2}{R} - W \sin{\theta}$$
##\frac{mv^2}{R}## must be greater than or equal to ##W \sin{\theta}## since tension can never be negative.
But if both terms are equal, then the tension of the string is zero. The particle no longer follows a circular path if ##T = 0##, since this is equivalent to the absence of a string holding on to the particle, right?
So the condition for pure circular motion (no coriolis or linear radial acceleration terms) should really be:
$$\frac{mv^2}{R} > W \sin{\theta}$$
without an equal sign, is this correct?