Chapter 11, Problem 41AP

Native people throughout North and South America used a bola to hunt for birds and animals. A bola can consist of three stones, each with mass m, at the ends of three light cords, each with length ℓ. The other ends of the cords are tied together to form a Y. The hunter holds one stone and swings the other two above his head (Figure P11.41a, page 308). Both these stones move together in a horizontal circle of radius 2ℓ with speed v₀. At a moment when the horizontal component of their velocity is directed toward the quarry, the hunter releases the stone in his hand. As the bola flies through the air, the cords quickly take a stable arrangement with constant 120-degree angles between them (Fig. P11.41b). In the vertical direction, the bola is in free fall. Gravitational forces exerted by the Earth make the junction of the cords move with the downward acceleration $\stackrel{\to }{g}$. You may ignore the vertical motion as you proceed to describe the horizontal motion of the bola. In terms of m, ℓ, and v₀, calculate (a) the magnitude of the momentum of the bola at the moment of release and, after release, (b) the horizontal speed of the center of mass of the bola, and (c) the angular momentum of the bola about its center of mass. (d) Find the angular speed of the bola about its center of mass after it has settled into its Y shape. Calculate the kinetic energy of the bola (e) at the instant of release and (f) in its stable Y shape. (g) Explain how the conservation laws apply to the bola as its configuration changes. Robert Beichner suggested the idea for this problem.

Figure P11.41

To determine

The magnitude of momentum of the bola at the moment of release and after the release.

Answer to Problem 41AP

The magnitude of momentum of the bola at the moment of release and after the release is $2m{v}_0$.

Explanation of Solution

At the moment of release, two stones are moving with speed $v_0$.

The total momentum of the is system is written as,

    $\begin{array}{c}p=m{v}_0+m{v}_0\\ =2m{v}_0\end{array}$

The total momentum has magnitude of $2m{v}_0$. It keeps this horizontal component of the momentum as it flies away.

Conclusion:

Therefore, the magnitude of momentum of the bola at the moment of release and after the release is $2m{v}_0$.

To determine

The horizontal speed of the centre of mass of the bola.

Answer to Problem 41AP

The horizontal speed of the centre of mass of the bola is $\frac{2v_0}{3}$.

Explanation of Solution

The centre of mass speed relative to the hunter is,

    $v_{\text{CM}}=\frac{p}{M}$

Mass of each stone is $m$, so mass of three stones will be $3m$.

Substitute $2m{v}_0$ for $p$ and $3m$ for $M$ in above equation.

    $\begin{array}{c}{v}_{\text{CM}}=\frac{2m{v}_0}{3m}\\ =\frac{2v_0}{3}\end{array}$

Conclusion:

Therefore, the horizontal speed of the centre of mass of the bola is $\frac{2v_0}{3}$.

To determine

The angular momentum of the bola about its centre of mass.

Answer to Problem 41AP

The angular momentum of the bola about its centre of mass is $\frac{4ml{v}_0}{3}$.

Explanation of Solution

The mass of each stone is $m$, the length of chord is $l$, the radius of horizontal circle is $2l$, the speed is $v_0$, the angle is $120°$ and the downward acceleration is $\overrightarrow{\text{g}}$.

When the bola is first released, the stones are horizontally in line with two at distance $l$ on one side of the centre knot and one at distance $l$ on the other side.

The centre of mass is given as,

    $\begin{array}{c}{x}_{\text{CM}}=\frac{2ml-ml}{3m}\\ =\frac{ml}{3m}\\ =\frac{l}{3}\end{array}$

This distance from the centre is not closer to the two stones: the one stone just being released at distance $r_1=\frac{4l}{3}$ from the centre of mass, the other two stones are released at distance $r_2=\frac{2l}{3}$.

The relative speed of the two stones is calculated as,

    $\begin{array}{c}{v}_2=v_0-\frac{2v_0}{3}\\ =\frac{v_0}{3}\end{array}$

The relative speed of first stone with respect to centre of mass is given as,

  $\begin{array}{c}{v}_1=\frac{2v_0}{3}-0\\ =\frac{2v_0}{3}\end{array}$

The angular speed of stone 1 is,

    ${\omega }_1=\frac{v_1}{r_1}$

Substitute $\frac{2v_0}{3}$ for $v_1$ and $\frac{4l}{3}$ for $r_1$ in above equation to find ${\omega }_1$.

    $\begin{array}{c}{\omega }_1=\frac{\frac{2v_0}{3}}{\frac{4l}{3}}\\ =\frac{v_0}{2l}\end{array}$

The angular speed of other two stones is,

    ${\omega }_2=\frac{v_2}{r_2}$

Substitute $\frac{v_0}{3}$ for $v_2$ and $\frac{2l}{3}$ for $r_2$ in above equation to find ${\omega }_2$.

    $\begin{array}{c}{\omega }_2=\frac{\frac{v_0}{3}}{\frac{2l}{3}}\\ =\frac{v_0}{2l}\end{array}$

The angular speed of other two stones is equal to the angular speed of stone 1.

    $\begin{array}{c}{\omega }_1={\omega }_2\\ =\omega \end{array}$

The total angular momentum is,

    $\begin{array}{c}L={\displaystyle \sum mvr}\\ =m{v}_1{r}_1+2m{v}_2{r}_2\end{array}$

Substitute $\frac{v_0}{3}$ for $v_2$, $\frac{2l}{3}$ for $r_2$, $\frac{2v_0}{3}$ for $v_1$ and $\frac{4l}{3}$ for $r_1$ in above equation to find $L$.

    $\begin{array}{c}L=m\left(\frac{2v_0}{3}\right)\left(\frac{4l}{3}\right)+2m\left(\frac{v_0}{3}\right)\left(\frac{2l}{3}\right)\\ =\frac{4ml{v}_0}{3}\end{array}$

Conclusion:

Therefore, the angular momentum of the bola about its centre of mass is $\frac{4ml{v}_0}{3}$.

To determine

The angular speed of the bola about its centre of mass after it has settled into its Y shape.

Answer to Problem 41AP

The angular speed of the bola about its centre of mass after it has settled into its Y shape is $\frac{4v_0}{9l}$.

Explanation of Solution

As the calculation of part (c), the angular speed $\omega$ at the moment of release is $\frac{v_0}{2l}$. As it moves through the air, the bola keeps constant angular momentum, but its moment of inertia changes to $3m{l}^2$.

The angular momentum is given as,

    $\begin{array}{c}L=I\omega \\ \frac{4ml{v}_0}{3}=3m{l}^2\omega \end{array}$

Rearrange the above expression for $\omega$.

    $\begin{array}{c}\omega =\frac{\frac{4ml{v}_0}{3}}{3m{l}^2}\\ =\frac{4v_0}{9l}\end{array}$

Conclusion:

Therefore, the angular speed of the bola about its centre of mass after it has settled into its Y shape is $\frac{4v_0}{9l}$.

To determine

The kinetic energy of the bola at the instant of release.

Answer to Problem 41AP

The kinetic energy of the bola at the instant of release is $m{v}_{\text{0}}^2$.

Explanation of Solution

The formula to calculate kinetic energy of the system is,

    $\begin{array}{c}KE=\frac{1}{2}m{\left(0\right)}^2+\frac{1}{2}\left(2m\right){\left(v_0\right)}^2\\ =m{v}_{\text{0}}^2\end{array}$

Conclusion:

Therefore, the kinetic energy of the bola at the instant of release is $m{v}_{\text{0}}^2$.

To determine

The kinetic energy of the bola in its stable Y shape.

Answer to Problem 41AP

The kinetic energy of the bola in its stable Y shape is $\frac{26}{27}m{v}_{\text{0}}^2$.

Explanation of Solution

The formula to calculate kinetic energy of the system is,

    $KE=\frac{1}{2}\left(3m\right){\left(v_{\text{CM}}\right)}^2+\frac{1}{2}I{\omega }^2$

Substitute $\frac{2v_0}{3}$ for $v_{\text{CM}}$, $3m{l}^2$ for $I$ and $\frac{4v_0}{9l}$ for $\omega$ in above equation.

    $\begin{array}{c}KE=\frac{1}{2}\left(3m\right){\left(\frac{2v_0}{3}\right)}^2+\frac{1}{2}\left(3m{l}^2\right){\left(\frac{4v_0}{9l}\right)}^2\\ =\frac{26}{27}m{v}_{\text{0}}^2\end{array}$

Conclusion:

Therefore, the kinetic energy of the bola in its stable Y shape is $\frac{26}{27}m{v}_{\text{0}}^2$.

To determine

The application of the conservation laws to the bola as its configuration changes.

Answer to Problem 41AP

The conservation laws are applied to the bola as it transforms its mechanical energy in to the internal energy to come in the stable state.

Explanation of Solution

The conservation laws states that the certain physical properties do not change in the course of time within an isolated physical system. There is no horizontal force act on the bola from the outside after release, so the horizontal momentum stays constant. Its center of mass moves steadily with the horizontal velocity it had at release.

No torques about its axis of rotation act on the bola, so its spin angular momentum stays constant. Internal forces cannot affect momentum conservation and angular momentum conservation, but they can affect mechanical energy. The cords pull on the stones as the stones rearrange themselves, so the cords must stretch slightly, so that energy $\frac{m{v}_0^2}{27}$ changes from mechanical energy into the internal energy as the bola takes its configuration.

Conclusion:

Therefore, the conservation laws are applied to the bola as it transforms its mechanical energy in to the internal energy to come in the stable state