Question

Why is the following situation impossible? The object of mass $m=4.00kg$ in below figure is attached to a vertical rod by two strings of length, $l=2.00m$. The strings are attached to the rod at points a distance $d=3.00m$ apart. The object rotates in a horizontal circle at a constant speed of $v=3.00m/s$, and the strings remain taut. The rod rotates along with the object so that the strings do not wrap onto the rod. What If? Could this situation be possible on another planet.

Answer 1

We solve for the tension in the two strings:
$F_g=mg=\left(4.00kg\right)\left(9.80m/s^2\right)=39.2N$
The angle $\theta$ is given by
$\theta ={\sin }^{-1}\left(\frac{1.50m}{2.00m}\right)={48.6}^0$
The radius of the circle is then
$r=\left(2.00m\right)\cos {48.6}^0=1.32m$
Applying Newton's second law,
$\sum F_x=m{a}_x=\frac{m{v}^2}{r}$
$T_a\cos {48.6}^0+T_b\cos {48.6}^0=\frac{\left(4.00kg\right)(3.00m/s{)}^2}{1.32m}$
$T_a+T_b=\frac{27.27N}{\cos {48.6}^0}=41.2N$............[1]
$\sum F_y=m{a}_y:T_a\sin {48.6}^0-T_b\sin {48.6}^0=39.2N=0$
$T_a-T_b=\frac{39.2N}{\sin {48.6}^0}=52.3N$............[2]
To solve simultaneously, we add the equation in $T_a$ and $T_b$:
$(T_a+T_b)+(T_a-T_b)=41.2N+52.3N$
$T_a=\frac{93.8N}{2}=46.9N$
This means that $T_b=41.2N–T_a=–5.7N$, which we may interpret as meaning the lower string pushes rather than pulls!
The situation is impossible because the speed of the object is too small, requiring that the lower string act like a rod and push rather than like a string and pull.
To answer the What if?, we go back to equation [2] above and substitute $mg$ for the weight of the object. Then,
$\sum F_y=m{a}_y:T_a\sin {48.6}^0-T_b\sin {48.6}^0-mg=0$
$T_a-T_b=\frac{\left(4.00kg\right)g}{\sin {48.6}^0}=5.33g$
We then add this equation to equation [2] to obtain
$(T_a+T_b)+(T_a+T_b)=41.2N+5.33g$
or $T_a=20.6N+2.67g$ and $T_b=41.2N-T_a=41.2N-2.67g$
For this situation to be possible, $T_b$ must be $>0$, or $g<7.72m/s^2$. This is certainly the case on the surface of the Moon and on Mars.