Question

Collision and rotation problems?

A cube with mass `M` and length `2a`, move in the velocity `v` on the friction-less horizontal table, when it closes to the end of the table, it is stopped by the long block(the height can be ignored), ask what is the value of `v` to topple from the table? ( note: use one side of cube as the rotating axis and moment of inertia is `8Ma^2/3`

Answer 1

Just after the collision the cube comes to stop. Hence, it has zero velocity. By Law of Conservation of Energy, all its translational kinetic energy gets converted into its rotational kinetic energy.

Rotational KE`=1/2I_"side"omega^2` .........(1)
where `I` is moment of inertia around edge (axis of rotation) and `omega` is the angular velocity of rotating cube.

When the cube is just about to topple over this rotational kinetic energy becomes zero and gets converted to its potential energy.

Initially, the center of mass of the cube is at a height `a` from the top of the table. The length of face diagonal of the cube is `=2sqrt2a`. When the cube is about to topple over, the center of mass is raised to a height `=sqrt2a`

Change in PE`=Mg(sqrt2a-a)` .......(2)
where `g` is acceleration due to gravity.

Equating (1) and (2) we get

`1/2I_"side"omega^2=Mg(sqrt2a-a)` ........(3)

`I_"side"=8/3Ma^2`, given. To find `omega` we make use of Law of Conservation of angular momentum.

Initial angular momentum`=` Final angular momentum
`Mvecvxxvecr=I_"side"omega`
`=>Mv_mina=I_"side"omega` ...........(4)
`=>Mv_mina=8/3Ma^2omega`
`=> omega=(3v_min)/(8a)` ........(5)

Inserting values from (4) and (5) in (3) we get

`1/2Mv_mina(3v_min)/(8a)=Mg(sqrt2a-a)`
`=>v_min^2=16/3ga(sqrt2-1)`
`=>v_min=4sqrt((ga)/3(sqrt2-1))`