Angular momentum of a rigid body physics problem

Question

I came across this task on a website and have been curious about its solution, but I’m having trouble figuring it out:

A cube with side edges d=2m, with homogeneous mass distribution, slides with speed v on horizontal, frictionless ground. All of a sudden, the cube reaches the slope connected to the ground, with an inclination angle of β=30º. The "break line" of the slope and the ground is perpendicular to the cube's direction of travel. The edge of the first side of the cube that is in contact with the ground stops perfectly inelastically at the fracture line, so the cube tips over. What is the minimum value of v if the front face of the cube "tips" on the slope?

Thanks in advance for your help.

Answer 1

Hi Leo,

It looks like the problem can be solved by conservation of energy, and it does not depends on the size of the angle of incline as long as it is less than 45°.

In in order for the cube to fall on the incline it's Center of gravity must be able to reach its highest point. This highest point is when the cube is standing on its edge with angles the two faces that form this edge are making equal angles to the horizontal, which is 45°.

Now, at the point the cube starts tipping, it's Center of mass is the height which is 1/2 of the length of its edge, which is 1 m. When the cube reaches the highest position, it's Central Mass as can be seen is elevated at the height equals to half of its edge length times square of 2.

Therefore potential energy of the cube will be raised from

PE=Mg*1m to PE=Mg*1m*Sqrt(2) where M is the cube's mass, and g - acceleration of gravity

In order to afford such increase of potential energy, the cube must have kinetic energy

KE=MV^2/2

at start which is equal to that

KE=MV^2/2 = ΔPE=Mg1m(Sqrt(2)-1)

From here find the minimum speed necessary :

V^2=2g(Sqrt(2)-1) and V=Sqrt[2g(Sqrt(2)-1)]

After the qube goes through the highest point, it could then fall on any incline with angle less then 45°

Best

Dr.Ariel B.