Chapter 5, Problem 18P

Why is the following situation impossible? The object of mass m = 4.00 kg in Figure P5.18 is attached to a vertical rod by two strings of length ℓ = 2.00 m. The strings are attached to the rod at points a distance d = 3.00 m apart. The object rotates in a horizontal circle at a constant speed of v = 3.00 m/s, and the strings remain taut. The rod rotates along with the object so that the strings do not wrap on to the rod. What If? Could this situation be possible on another planet?

To determine

The reason why the situation shown in Figure P5.18 is impossible, and whether this situation is be possible on another planet.

Answer to Problem 18P

The situation shown in Figure P5.18 is impossible, because the speed of the object is too small, the lower string is require that act like a rod and push rather than like a string and pull. This situation is only possible when $T_{\text{b}}$ must be greater than zero or $g<7.72\text{m}/{\text{s}}^2$ on the other planet.

Explanation of Solution

The free body diagram of the system is shown Figure.

Write the expression for force due to gravity.

    $F_{\text{g}}=mg$        (I)

Here, $F_{\text{g}}$ is the force due to gravity, $m$ is the mass, and $g$ is the acceleration due to gravity.

A centripetal force is needed to keep the object in the circular motion, this is equal to the force in the horizontal direction.

    ${\displaystyle \sum F_x}=\frac{m{\upsilon }^2}{r}$        (II)

Here, $\upsilon$ is the speed, and $r$ is the radius of the object.

From the free body diagram, write the expression for net force in the $x$ direction.

    ${\displaystyle \sum F_x}=T_{\text{a}}\cos \theta +T_{\text{b}}\cos \theta$        (III)

Here, $T_{\text{a}}$ is the tension towards upward string, $T_{\text{b}}$ is the tension towards downward, and $\theta$ is the angle.

Equate equation (II) and (III).

    $\begin{array}{c}{T}_{\text{a}}\cos \theta +T_{\text{b}}\cos \theta =\frac{m{\upsilon }^2}{r}\\ T_{\text{a}}+T_{\text{b}}=\frac{m{\upsilon }^2}{r\cos \theta }\end{array}$        (IV)

Write the expression for force in the $y$ direction.

    ${\displaystyle \sum F_y}=0$        (V)

Since there is no acceleration in the $y$ direction, the net force is zero.

From the free body diagram, write the expression for net force in the $y$ direction.

    ${\displaystyle \sum F_y}=T_{\text{a}}\cos \theta +T_{\text{b}}\cos \theta -F_{\text{g}}$        (VI)

Equate equation (V) and (VI).

    $\begin{array}{c}{T}_{\text{a}}\sin \theta -T_{\text{b}}\sin \theta -F_{\text{g}}=0\\ T_{\text{a}}-T_{\text{b}}=\frac{F_{\text{g}}}{\sin \theta }\end{array}$        (VII)

Add equation (IV) and (VII).

    $\begin{array}{c}{T}_{\text{a}}+T_{\text{b}}+T_{\text{a}}-T_{\text{b}}=\frac{m{\upsilon }^2}{r\cos \theta }+\frac{F_{\text{g}}}{\sin \theta }\\ 2T_{\text{a}}=\frac{m{\upsilon }^2}{r\cos \theta }+\frac{F_{\text{g}}}{\sin \theta }\\ T_{\text{a}}=\frac{1}{2}\left(\frac{m{\upsilon }^2}{r\cos \theta }+\frac{F_{\text{g}}}{\sin \theta }\right)\end{array}$        (VII)

Conclusion:

The angle $\theta$ can be found from the figure P5.18 as follows,

`    $\begin{array}{c}\sin \theta =\frac{15.2\text{m}}{2\text{m}}\\ \theta =48.6°\end{array}$

The radius of the orbit can be found as follows,

    $\begin{array}{c}\cos \theta =\frac{r}{2\text{m}}\\ r=2\text{m}\left(\cos 48.6°\right)\\ =1.32\text{m}\end{array}$

Substitute, $4\text{kg}$ for $m$, $3\text{m}/\text{s}$ for $\upsilon$, $1.32\text{m}$ for $r$, $48.6°$ for $\theta$, $\left(4\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)$ for $F_{\text{g}}$ in the equation (VII), to find $T_{\text{a}}$.

    $\begin{array}{c}{T}_{\text{a}}=\frac{1}{2}\left[\frac{\left(4\text{kg}\right){\left(3\text{m}/\text{s}\right)}^2}{\left(1.32\text{m}\right)\cos \left(48.6°\right)}+\frac{\left(4\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)}{\sin \left(48.6°\right)}\right]\\ =\frac{1}{2}\left(41.2\text{N+52}\text{.3}\text{N}\right)\\ =46.9\text{N}\end{array}$

Substitute, $46.9\text{N}$ for $T_{\text{a}}$, $4\text{kg}$ for $m$, $3\text{m}/\text{s}$ for $\upsilon$, $1.32\text{m}$ for $r$, and $48.6°$ for $\theta$ in the equation (IV) for $T_{\text{b}}$.

    $\begin{array}{c}46.9\text{N}+T_{\text{b}}=\frac{\left(4\text{kg}\right){\left(3\text{m}/\text{s}\right)}^2}{\left(1.32\text{m}\right)\cos \left(48.6°\right)}\\ =41.2\text{N}\\ T_{\text{b}}=41.2\text{N}-46.9\text{N}\\ =-5.7\text{N}\end{array}$

It indicates that lower string pushes rather than pulls.

Substitute, $4\text{kg}$ for $m$, $3\text{m}/\text{s}$ for $\upsilon$, $1.32\text{m}$ for $r$, $48.6°$ for $\theta$, $\left(4\text{kg}\right)g$ for $F_{\text{g}}$ in the equation (VII), to find $T_{\text{a}}$.

    $\begin{array}{c}{T}_{\text{a}}=\frac{1}{2}\left[\frac{\left(4\text{kg}\right){\left(3\text{m}/\text{s}\right)}^2}{\left(1.32\text{m}\right)\cos \left(48.6°\right)}+\frac{\left(4\text{kg}\right)g}{\sin \left(48.6°\right)}\right]\\ =\frac{1}{2}\left(41.2\text{N+5}\text{.33g}\right)\\ =20.6\text{N+2}\text{.67g}\end{array}$

Substitute, $20.6\text{N+2}\text{.67g}$ for $T_{\text{a}}$, $4\text{kg}$ for $m$, $3\text{m}/\text{s}$ for $\upsilon$, $1.32\text{m}$ for $r$, and $48.6°$ for $\theta$ in the equation (IV) for $T_{\text{b}}$.

    $\begin{array}{c}20.6\text{N+2}\text{.67g}+T_{\text{b}}=\frac{\left(4\text{kg}\right){\left(3\text{m}/\text{s}\right)}^2}{\left(1.32\text{m}\right)\cos \left(48.6°\right)}\\ =41.2\text{N}\\ T_{\text{b}}=41.2\text{N}-20.6\text{N}-\text{2}\text{.67g}\\ =20.6\text{N+2}\text{.67g}\end{array}$

This is possible only when $T_{\text{b}}>0$ or $g<7.72\text{m}/{\text{s}}^2$.

Therefore, the situation shown in Figure P5.18 is impossible, because the speed of the object is too small, the lower string is require that act like a rod and push rather than like a string and pull. This situation is only possible when $T_{\text{b}}$ must be greater than zero or $g<7.72\text{m}/{\text{s}}^2$ on the other planet.