Question 11.8: A cube with the edge length 2a and mass M glides with consta...
A cube with the edge length 2a and mass M glides with constant velocity v_{0} on a frictionless plate. At the end of the plate, it bumps against an obstacle and tilts over the edge (see Fig. 11.15). Find the minimum velocity v_{0} for which the cube still falls from the plate!
We look for the velocity v_{0} for which the cube can tilt over its edge, as is represented in (c). If it bumps into the obstacle at the edge of the plate, it is set into rotation about the axis A. At the time of collision all external forces act along this axis, and the angular momentum of the cube is conserved. Before hitting the obstacle, the cube has—due to the translational motion—the angular momentum
L = |r×p| = p · a = Mv_{0}a. (11.19)
Immediately after the collision, the angular momentum appears as rotational motion of the cube
L = \Theta_{A}ω_{0} = Mv_{0}a,or
ω_{0} = \frac{Mv_{0}a}{\Theta_{A}}. (11.20)
If the cube begins to lift off, the gravitational force causes a torque about the axis A that counteracts the lifting process.
For the kinetic energy of the cube immediately after the collision, one has, for given ω_{0},
T_{0} = \frac{1}{2}\Theta_{A}ω^{2}_{0}= \frac{1}{2} \frac{M^{2}v^{2}_{0}a^{2}}{\Theta_{A}}. (11.21)
The potential energy difference between position a and position c is
ΔV = M(h_{2} −h_{1})g = M(\sqrt{2}a −a)g = Mag(\sqrt{2} −1), (11.22)
and from the energy conservation law, immediately it follows that
Mag(\sqrt{2}− 1) = \frac{1}{2}\frac{M^{2}v^{2}_{0}a^{2}}{\Theta_{A}}. (11.23)
The moment of inertia of the cube \Theta_{A} is easy to calculate:
\Theta_{A}= \varrho \int\limits_{V}{r^{2} dV }= \varrho \int\limits_{0}^{2a}{\int\limits_{0}^{2a}{\int\limits_{0}^{2a}{(x^{2} + y^{2})dx dy dz}}} = \frac{8}{3}Ma^{2}. (11.24)
From (11.23), it follows that
ag(\sqrt{2}− 1) = \frac{1}{2} \frac{Ma^{2}}{(8/3)Ma^{2}}v^{2}_{0},
and from this, we find
v_{0} =\sqrt{ag \frac{16}{3}(\sqrt{2}− 1)}. (11.25)
This is the correct result.We emphasize this because one could easily come to another result by a false consideration: The kinetic energy is (1/2)Mv^{2}_{0} , and from the energy conservation combined with (11.22) it follows that
\frac{1}{2}Mv^{2}_{0}= Mag(\sqrt{2}− 1).This leads to
v_{0} =\sqrt{2ag(\sqrt{2}−1)}, (11.26)
i.e., a value that is smaller than the correct result (11.25) by the factor \sqrt{3/8}. The result (11.26) is wrong since the cube loses part of its kinetic energy in the collision because of its inelasticity. The correct result (11.25) is based upon the conservation of angular momentum, which acts “more strongly” than the conservation of energy.