Question

A 50 kg acrobat is swinging on a rope with a length of 15m from one horizontal platform to another. Both platforms are at equal height.
If the maximum tension the rope can support is 1200 N , which of the following answer best represents the maxim velocity the acrobat can reach without breaking the rope?
A. $25m/s$
B. $19m/s$
C. $20m/s$
D. $15m/s$
E. $5m/s$

Answer 1

Hint Whenever a body is moving in vertical circular motion the tension in the rope is different at different positions. And tension in rope is minimum at top most point where speed is slowest and maximum at bottom where speed is fastest. Tension at the lowermost point can be balanced by tension with centripetal force and gravitational force of the acrobat.

Complete step-by-step solution :As we’ve read that tension is maximum at bottom and minimum at top most point of the circular motion.
At the top most point of the circular motion centripetal force will act upwards and as usual gravitational force acts downwards. So we can give the tension at top most point as:
$T = \dfrac{{m{v^2}}}{r} - mg$
At the bottom most points of the circular point centripetal force and gravitational force both act downward so we can find the tension in the rope by adding these two quantities.
$T = \dfrac{{m{v^2}}}{r} + mg$
We are asked to find velocity at that point where velocity is maximum and so tension is; which is at the bottom most point of the circular motion.
Therefore,
$\dfrac{{m{v^2}}}{r} = T - mg$
We are given:
Mass $m = 50kg$
Length of the rope $r = 15m$
Maximum tension that rope can take is $T = 1200N$
Putting values in above equation:
$\dfrac{{50 \times {v^2}}}{{15}} = 1200 - 50 \times 10$
$\dfrac{{50 \times {v^2}}}{{15}} = 700$
${v^2} = \dfrac{{700 \times 15}}{{50}}$
$v = \sqrt {210}$
$v = 14.49m/s$
Which is approximately
$v = 15m/s$
Hence, option D is correct.

Note:- This is the case when circular motion is done through rope.
The minimum velocity needed(at bottom most point) to complete rotation is $\sqrt {5gl}$ in the same case minimum velocity which is at top most point is $\sqrt {gl}$ . When circular motion is done through rod maximum and minimum velocity is $\sqrt {4gl}$ and zero respectively.