Toss those cabers, toss them high

Let the mass of the caber be $m$ . Let the linear velocity of the caber be $v$ , and let the rotational velocity be $\omega$ . If $I$ is the rotational inertia of the caber, the kinetic energy will be: $K.E= \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2$ Let the length of the caber be $l$ . Modelling the caber as a thin rod, the moment of inertia of the caber will then be $\frac{ml^2}{12}$ (see this page for the list of moments of inertia). We have to minimize the kinetic energy, i.e we have to minimize $K.E= \frac{1}{2} mv^2 + \frac{ml^2}{24} \omega^2 \text{ .....(i)}$ Let the total time taken for the caber toss be $t$ , so after $t$ seconds the caber rotates $\pi$ . We then have: $t = \frac{\pi}{\omega }$
Note that after $t$ seconds the net vertical displacement of the caber is zero, so: $vt + \frac{1}{2} gt^2= 0$ $\implies v+\frac{1}{2}gt= 0$ $\implies v= \frac{-gt}{2}$ Plugging in the expression for $t$ ,we obtain: $v= -\frac{g \pi }{2 \omega }$ Plugging this value in $(i)$ , we obtain: $K.E= \frac{1}{2} m \left ( \frac{g \pi }{2 \omega } \right )^2 + \frac{ml^2}{24} \omega^2$ We wish to minimize $K.E$ . Applying AM-GM inequality, we obtain: $\frac{1}{2} m \left ( \frac{g \pi }{2 \omega } \right )^2 + \frac{ml^2}{24} \omega^2 \geq 2 \sqrt{\frac{1}{2} m \left ( \frac{g \pi }{2 \omega } \right )^2 \times \frac{ml^2}{24} \omega^2}$ $\implies K.E \geq \frac{\pi }{4 \sqrt{3}} mlg$ Equality holds iff $\frac{1}{2} m \left ( \frac{g \pi }{2 \omega } \right )^2 = \frac{ml^2}{24} \omega^2$ $\implies \omega = \sqrt{ \frac{g \pi}{2\sqrt{3}}}$ Plugging the provided values in the expression for minimum value of $K.E$ , we obtain $K.E \approx \boxed{2133 \ J}$ .

The movement executed by the caber will be a combination of a vertical launch (center of mass translation) and rotational movement around its own center of mass on a fixed plane. Thus, the tosser will give the caber both translational and rotational kinetic energy:

$K = K_T + K_R = \frac{Mv_{0}^2}{2} + \frac{I \omega^2}{2}$

Since the toss is vertically launched, the total time between the launch and landing is:

$t = \frac{2v_0}{g}$

During that time, the minimum angular displacement of the caber is 180 degrees (it could be 180° plus a multiple of 360°), as long as the other end of the caber must hit the ground. Since the only force acting on the rod is its own weight (and gravity is a central force), there are no external torques acting on the caber. Thus, its angular speed will be constant and equal to:

$\omega = \frac{\pi}{t} = \frac{\pi g}{2v_0}$

So the total kinetic energy of the caber is:

$K = \frac{Mv_{0}^2}{2} + \frac{I \omega^2}{2} = \frac{Mv_{0}^2}{2} + \frac{I \pi ^2 g^2}{8v_{0}^2}$

The best lower bound for $K$ can be found using the AM-GM inequality (both terms are positive):

$\frac{Mv_{0}^2}{2} + \frac{I \pi ^2 g^2}{8v_{0}^2} \geq 2\sqrt{\frac{Mv_{0}^2}{2} \frac{I \pi ^2 g^2}{8v_{0}^2} } = \frac{\pi g \sqrt{MI}}{2}$

Since the caber can be modeled as an uniform rod, $I = \frac{ML^2}{12}$ . Plugging that into the kinetic energy expression:

$K \geq \frac{\pi}{4 \sqrt{3}} MgL$ , for all possible values of $v_0$ .

Therefore, the minimum kinetic energy that must be given to the caber is

$K_{min} = \frac{\pi}{4 \sqrt{3}} MgL = \frac{\pi}{4 \sqrt{3}} \times 80 \times 9.8 \times 6= \boxed{2133.03 J}$

Let $\ell$ and $m$ be the length and mass of the cabre respectively.

The cabre when shot vertically takes time $t=\frac{2v_0}{g}$ to reach back to the ground. In this time, the cabre must rotate by an angle of $(2n+1) \pi$ where $\displaystyle n=0,1,2...$ so that the other end touches the ground. As we need the minimum kinetic energy, we set $n=0$ . The time taken by the cabre to rotate an angle of $\pi$ is $\frac{\pi}{\omega}$ , where $\omega$ is the angular velocity of the cabre. We can find the angular velocity by comparing the two expressions we have obtained for time, i.e.

$\displaystyle \frac{2v_0}{g}=\frac{\pi}{\omega} \Rightarrow \omega = \frac{\pi g}{2v_0}$

The expression for the initial kinetic energy of the cabre is, $\displaystyle K=\frac{1}{2}mv_0^2+ \frac{1}{2}I\omega^2$

where I is the moment of inertia of the cabre.

$\displaystyle K=\frac{1}{2}mv_0^2+\frac{1}{2} \cdot \frac{m \ell^2}{12} \cdot \frac{\pi^2 g^2}{4v_0^2}$

$\displaystyle \Rightarrow K=\frac{1}{2}m \left(v_0^2+\frac{\pi^2 g^2 \ell^2}{48v_0^2} \right)$

To find the minimum kinetic energy, differentiate the expression with respect $v_0$ and set the derivative equal to zero.

$\displaystyle \frac{dK}{dv_0}=\frac{1}{2}m \left(2v_0-\frac{\pi^2 g^2 \ell^2}{24v_0^3} \right) =0$

$\displaystyle \Rightarrow v_0^4=\frac{\pi^2 g^2 \ell^2}{48} \Rightarrow v_0^2=\frac{\pi g \ell}{\sqrt{48}}$

The expression for kinetic energy can be rewritten as

$\displaystyle \Rightarrow K=\frac{1}{2}m \left( \frac{48v_0^4+\pi^2 g^2 \ell^2}{48v_0^2} \right)$

Substituting the values of $v_0^4$ and $v_0^2$ ,

$\displaystyle K_{min}=\frac{1}{2}m \left( \cfrac{\pi^2 g^2 \ell^2+\pi^2 g^2 \ell^2}{48 \cfrac{\pi g \ell}{\sqrt{48}}} \right)$

$\displaystyle K_{min}=\frac{1}{2}m \cdot \frac{2 \pi^2 g^2 \ell^2}{\sqrt{48}\pi g \ell}$

Substituting the values of $m, g$ and $\ell$ , $K_{min}=2133.03 J$

Despite the rotation of the caber, its center of mass will move as a vertical projectile. We can find its time of flight using kinematics and analyzing the apex of flight: $v_f = v_0 + at_1$ $t_1 = \frac{v_f - v_0}{a} = \frac{0 - v_0}{-g} = \frac{v_0}{g}$ Now $t = 2t_1 = \frac{2v_0}{g}$ , since the time taken to reach the apex is half the total time of flight. In order for the caber to rotate a full 180 degrees ( $\pi$ radians), its angular velocity must be $\omega = \frac{\Delta\theta}{t} = \frac{\pi}{\frac{2v_0}{g}} = \frac{\pi g}{2v_0}$ . We can express the total kinetic energy given to the caber in terms of linear and rotational components: $\Sigma KE = KE_{linear} + KE_{rot} = \frac{1}{2}mv_0^2 + \frac{1}{2}I(\frac{\pi g}{2v_0})^2$ Since the caber can be modeled as a uniform rod rotating about its center, $I = \frac{1}{12}mL^2$ . Substituting into the KE expression: $KE = \frac{1}{2}mv_0^2 + \frac{1}{2}(\frac{1}{12}mL^2)(\frac{\pi^2 g^2}{4v_0^2}) = \frac{1}{2}mv_0^2 + \frac{\pi^2 L^2 g^2 m}{96v_0^2}$ Let this be a function of $v_0$ ; we now have to find the minimum value of $KE(v_0)$ . We can do this by finding the zero of the derivative: $\frac{dKE}{dv_0} = mv_0 - \frac{\pi^2 L^2 g^2 m}{48v_0^3} = 0$ Dividing by $mv_0$ : $1 - \frac{\pi^2 L^2 g^2}{48v_0^4} = 0$ $\frac{\pi^2 L^2 g^2}{48v_0^4} = 1$ $v_0 = \sqrt[4]{\frac{\pi^2 L^2 g^2}{48}} = \sqrt[4]{\frac{\pi^2(6)^2(9.8)^2}{48}} \approx 5.1636 m/s$ We now plug in to find the minimum kinetic energy: $KE_{min} = \frac{1}{2}(80)(5.1636)^2 + \frac{\pi^2(6)^2(9.8)^2(80)}{96(5.1636)^2} \approx\boxed{2133 J}$