Question

Two solid balls have different radii but are made of same material. The balls are linked together with a long thin thread and released from a large height. At the terminal velocity, the thread is under tension. The larger ball has a fixed mass, but we have choice of the smaller ball with different masses. At what ratio of larger and smaller mass will this tension be maximum?

Answer 1

The tension is maximum when $\frac{m_L}{m_S}\approx 4.74$.

Answer 2

We begin by noting that for a sphere moving through air the drag force is

$$F_D = \frac{1}{2}C_D \rho A v^2,$$

with cross–sectional area $A=\pi R^2$ and for a solid sphere of density $\rho_{\rm ball}$ the mass is

$$m=\frac{4}{3}\pi R^3\rho_{\rm ball}.$$

Thus if we eliminate $R$ we obtain

$$R\propto m^{1/3}\quad\Longrightarrow\quad A\propto m^{2/3}.$$

So we write the drag force in the form

$$F_D=\alpha\,m^{2/3}\,v^2,$$

where $\alpha$ is a constant (including $\frac{1}{2}C_D\rho$ and numerical factors).

Now we have two balls—the larger one has a fixed mass $m_L$ and the smaller one a mass $m_S$. When they are linked by a thread and falling at their common terminal speed $v$ (steady state) each ball obeys a force‐balance. For the larger ball (taking downward as positive, and noting that the drag is upward and the thread pulls upward) we have

$$m_Lg - F_{DL} - T=0\quad\Longrightarrow\quad T=m_Lg-\alpha\,m_L^{2/3}v^2,$$

while for the smaller ball (here the thread pulls downward on it) we obtain

$$m_Sg - F_{DS} + T=0\quad\Longrightarrow\quad T=\alpha\,m_S^{2/3}v^2-m_Sg.$$

Equate the two expressions for $T$:

$$m_Lg-\alpha\,m_L^{2/3}v^2=\alpha\,m_S^{2/3}v^2-m_Sg.$$

Solve for $v^2$:

$$\alpha\Bigl(m_S^{2/3}+m_L^{2/3}\Bigr)v^2=g\Bigl(m_L+m_S\Bigr),$$

so that

$$v^2=\frac{g(m_L+m_S)}{\alpha\left(m_S^{2/3}+m_L^{2/3}\right)}.$$

Substitute back (say, in the expression for the small ball) to get

$$T=\alpha\,m_S^{2/3}\,v^2 - m_Sg =\frac{g\,m_S^{2/3}(m_L+m_S)}{m_S^{2/3}+m_L^{2/3}}-m_Sg.$$

It is convenient to introduce the mass–ratio

$$r=\frac{m_L}{m_S}\quad(\text{with }r>1),$$

so that $m_L=r\,m_S$. Notice also that

$$m_L^{2/3}=(r\,m_S)^{2/3}=r^{2/3}m_S^{2/3}.$$

Then the numerator becomes

$$m_S^{2/3}(r\,m_S+m_S)= m_S^{5/3}(r+1),$$

and the denominator is

$$m_S^{2/3}\left(1+r^{2/3}\right).$$

Thus

$$T=\frac{g\,m_S^{5/3}\,(r+1)}{m_S^{2/3}\,(1+r^{2/3})}-m_Sg =\frac{g\,m_S\,(r+1)}{1+r^{2/3}}-m_Sg.$$

So we may write

$$T=g\,m_S\left[\frac{r+1}{1+r^{2/3}}-1\right] =\frac{g\,m_S\Bigl[(r+1) - (1+r^{2/3})\Bigr]}{1+r^{2/3}} =\frac{g\,m_S\left(r-r^{2/3}\right)}{1+r^{2/3}}.$$

It is now clear that for a fixed larger mass $m_L$ the smaller mass $m_S$ is a free parameter. Writing

$$m_S=\frac{m_L}{r},$$

the tension becomes

$$T=\frac{g\,m_L}{r}\cdot\frac{r-r^{2/3}}{1+r^{2/3}} =g\,m_L\,\frac{r^{1/3}-1}{1+r^{2/3}}\,r^{-1/3}.$$

Let

$$x=r^{1/3}\quad\Longrightarrow\quad r=x^3,$$

so that

$$T=g\,m_L\,\frac{x-1}{x(1+x^2)}.$$

Since $g$ and $m_L$ are constant, maximizing the tension is equivalent to maximizing

$$f(x)=\frac{x-1}{x(1+x^2)}$$

for $x>1$.

Differentiate $f(x)$:

$$f'(x)=\frac{d}{dx}\left[\frac{x-1}{x+x^3}\right].$$

Using the quotient rule (with $u=x-1$ and $v=x+x^3$; note $u'=1$ and $v'=1+3x^2$):

$$f'(x)=\frac{u'v - u\,v'}{v^2}=\frac{\bigl[1\cdot (x+x^3)\bigr] -\bigl[(x-1)(1+3x^2)\bigr]}{(x+x^3)^2}.$$

Expanding,

$$x+x^3-(x-1)(1+3x^2)=x+x^3-\bigl(x+3x^3-1-3x^2\bigr) =x+x^3-x-3x^3+1+3x^2 =1-2x^3+3x^2.$$

Setting the numerator equal to zero gives the equation

$$1-2x^3+3x^2=0\quad\Longrightarrow\quad 2x^3-3x^2-1=0.$$

This cubic in $x$ has a unique real solution for $x>1$. A numerical solution shows:

• At $x=1$: $2-3-1=-2$ (negative),
• At $x=1.7$: $2(4.913)-3(2.89)-1\approx 9.826-8.67-1\approx 0.156$ (close to 0).

Refining the estimate we find approximately

$$x\approx1.68.$$

Since $r=x^3$, the optimum mass–ratio is

$$r\approx (1.68)^3\approx 4.74.$$

Thus, the thread tension is maximized when the mass–ratio is

$$\frac{m_L}{m_S}\approx 4.74.$$

In summary: Write the drag force on each ball as proportional to $m^{2/3}v^2$, set up force–balances for the large and small balls, equate the tensions, substitute $r=m_L/m_S$ (with $m_S=m_L/r$) and express the tension in terms of $r$. Then by substituting $x=r^{1/3}$ the tension becomes proportional to $\frac{x-1}{x(1+x^2)}$. Differentiating with respect to $x$ and setting the derivative to zero gives $x\approx1.68$ so that $r\approx(1.68)^3\approx4.74$.