A solid cube of side $2a$ and mass $M$ (moment of inertia for a cube of side $s$ about its center of mass is $I=1/6 Ms^2$) is sliding on a frictionless surface with constant velocity. It hits a small obstacle (inelastic collision) at the end of the table that causes the cube to tilt over as shown. Show that the minimum speed that the cube needs to tip over and fall of the table is: (Hint: you’ll need to use energy here)

$\displaystyle v_{min}=4\sqrt{\frac{ga}{3}(\sqrt{2}-1)}$

enter image description here

My attempt:

For minimum initial velocity, the cube should have zero velocity when it is just about to tip over. So, the initial kinetic energy must be completely converted into potential energy. Initially, the cube's center of mass is at a height $a$ (measured from the top of the table). The length of a diagonal of a side of the cube is $2\sqrt 2a$. So, when the cube is about to tip over, the center of mass is at a height of $\sqrt 2a$.

$\frac{1}{2}mv^2=mg(\sqrt 2a-a)$

$v^2=2g(\sqrt 2a-a)$

$v=\sqrt{2ga(\sqrt 2-1)}$

This is not the answer given, so my approach is obviously wrong. Can someone tell me how to solve this problem?

  • $\begingroup$ At the moment of impact the velocity of the contact point is zero, and not of the center of mass (where you measure KE). $\endgroup$ – ja72 Feb 11 '17 at 2:07
  • $\begingroup$ Your mistake is that you have applied conservation of kinetic energy. This is wrong because the question tells you that the collision with the small obstacle is inelastic. But angular momentum is conserved. $\endgroup$ – sammy gerbil Feb 12 '17 at 18:33

You have made a conceptual mistake. Remember one thing in rotational dynamics - If there is some point on a rotating body which is not moving then that is the point around which the body is rotating.

In the above case the point of contact of the cube with the table when it is tipping over is that point as it is not moving ( there is no slipping ) .

So, the translational kinetic energy is converted into rotational kinetic energy just after the collision. And you need to equalise this Rotational kinetic energy to $mg(\sqrt 2a-a)$ and not the translational kinetic energy.

To find the initial angular velocity of the block, you need to conserve angular momentum about $A$. Yes, there will be a frictional force provided by the point $A$ but you see, the torque by that frictional force will be zero so you can conserve the angular momentum about $A$.

Here is a picture of the solution. I am sorry that I was not able to type the solution due to lack of time. enter image description here


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