How to account for air resistance in a sphere being released against a flow or air?

Question

The problem is as follows:

A bob is hanging from the ceiling of a specially designed room where a flow of air is being blown against. The bob is makes a $53^{\circ}$ angle with respect of the ground as indicated in the figure. The mass of the sphere is $2\,kg$. Assume that the wire is ideal. The flow of air excerts a constant force whose modulus is $4\,N$. Given these conditions, find the modulus of the force in $N$ in the wire when the sphere passes through its lowest point.

The alternatives are as follows:

$\begin{array}{ll} 1.&26.9\,N\\ 2.&29.6\,\frac{m}{s^2}\\ 3.&27.5\,\frac{m}{s^2}\\ 4.&23.3\,\frac{m}{s^2}\\ 5.&21.2\,\frac{m}{s^2}\\ \end{array}$

I'm not sure exactly how should I account for the force of air going against the bob.

What I think should be used here is the conservation of mechanical energy:

$E_u=E_k$

$mgh=\frac{1}{2}mv^2$

When the ball passes through the lowest point will be:

$T-mg=\frac{mv^2}{R}$

$T=\frac{mv^2}{R}+mg$

From the first equation:

$mgh=\frac{1}{2}mv^2$

$2g(1-\sin 53^{\circ})=\frac{v^2}{R}$

Therefore the tension will be:

$T=mg+\frac{mv^2}{R}=mg+2mg(1-\sin 53^{\circ})$

Therefore:

$T=2\times 10 + 2 \times 10 (1-\frac{4}{5})= 20 +20(\frac{1}{5})=24\,N.$ But this is not within the alternatives, needless to say that this doesn't seem to be the right answer. Can someone help here please?. I'm still stuck here.

Answer 1

Equations:

$l$: length of wire

1) $\Delta E_{pot}=l(1-\cos 53)mg$

2)$ \Delta E_{pot}=(1/2)mv^2+4l \sin 53$

3) $T-mg=mv^2/l$

Answer 2

Because of the peculiar definition of “air resistance” in this problem, namely the fact that the force of “air resistance” is independent of the velocity of the sphere, you have two choices how to handle it.

One way is to treat the “air resistance” like the air resistance in most other problems, and say that work the sphere does against the air resistance represents the transfer of the sphere’s mechanical energy into other energy.

In that interpretation, the mechanical energy of the sphere is not conserved. To see how much mechanical energy is lost between the starting position and the lowest point in the sphere’s path, compute the work done against a $-4$ N horizontal force along that part of the path. The mechanical energy is reduced by that amount.

Another way is to say that no matter where the sphere is or what it is doing, it always is subjected to a force in a direction downward and to the left, and that this force has components $-4$ in the $x$ direction and $-20$ in the $y$ direction. This is a conservative force, and you can plot lines of equal potential energy in the force field. But since the force is not straight down, these lines of equal potential are not horizontal.

According to this interpretation, the mechanical energy of the sphere is conserved, but the potential is not $mgy$. Instead, it’s a function of both $x$ and $y$.