Given:

Square box which is uniformly loaded is released on the floor when its corner D is directly above the A. square box tends to rotate till it strikes the floor. The floor is made anti-slipping and impact at B is perfectly plastic.

Concept:

According to impulse momentum principle,

Calculation:

Let’s consider,

M = mass of a square box.

C = length of the side.

Moment of inertia = I=1/2m(c2+c2)=1/6mc2

We can say,

v0=rG/Axω0=1/2ω0v=rG/Bxω=1/2ω

Taking moment at A,

Iω0+0=Iω+rG/Bx mv1/6mc2ω0=1/6mc2ω+(1/2)m(1/2)1/6mc2ω0=1/6mc2ω+[/4]1/6mc2ω0=1/6mc2ω+2c2mω/4mc2ω0/6=mc2ω/6+2c2mω/4mc2ω0/6=mc2ω/2(1/3+2/2)mc2ω0/6=4/1(mc2ω/2)3/4x2/6xmc2ω0/mc2=ω1/4ω0=ω

Conclusion:

The angular velocity of the square box after striking B on the floor will be 14 of the angular velocity before B strikes on the floor.

Given:

Square box which is uniformly loaded is released on the floor when its corner D is directly above the A. square box tends to rotate till it strikes the floor. The floor is made anti-slipping and impact at B is perfectly plastic.

Calculation:

Kinetic energy before impact,

E1=1/2 Iω02+1/2mv02=1/2(1/6mc2)ω02+1/2m(1/2cω0)2=mc2ω02/12+m/2(2/4c2ω02)=mc2ω02/12+mc2ω02/4=mc2ω02(1/12+1/4)E1=1/3 mc2ω02

Similarly kinetic energy after impact,

E2=1/2 Iω2+1/2mv2=1/2(1/6mc2)ω2+1/2m(1/2cm)2=mc2ω2/12+mc2ω2/4=1/3mc2ω2=1/3mc2(1/4ω0)2E2=1/48mc2ω02

Combining E₁ and E₂ to find fraction of energy cost,

E1−E2/E1=1/3mc2ω0– 1/48mc2ω0/1/3mc2ω0=(1/3−1/48)mc2ω02/(1/3)mc2ω02=1/3−1/48/1/3=(1/3/(1/3)−(1/48)/(1/3)=3/3−3/48=1−1/1615/16

Conclusion:

The fraction of energy lost during impact conditions is 1516.

Given:

Square box which is uniformly loaded is released on the floor when its corner D is directly above the A. square box tends to rotate till it strikes the floor. The floor is made anti-slipping and impact at B is perfectly plastic.

Concept:

As per law of conservation of energy before impact,

T0+V0=T1+V1

And that of after impact,

T3+V3=T2+V2T0=0,T3=0, T2=1/16Tv0=mg(1/22c)v1=v2=mg(1/22c)v3=mgh3

Before impact,

T0+V0=T1+V1T1=V0−V1=mg(1/22c)=mg(1/22c)−mg(1/2c)=mg2c/2−mgc/2T1=mgc/2(2−1)

After impact,

T3+V3=T2+V2T2=V3−V2=mgh3−mgc/2T2=mg(h3−c/2)

As,

T2=1/16T11/16mg c/2(2−1)=mg(h3−c/2)2−1/32c=h3−c/2h3=((2−1)c/32+c/2=((2−1)c+16c/32h3=((2−1)+16)c/32

But

Therefore, from geometry,

h3=1/22csin(θ+450)

Equating both equations of h₃

[(2−1)+16/32]c=2c/2sin(θ+450)[(2−1)+16/32]c × 2/2c=sin(θ+450)2−1/16+1=sin(θ+450)1.0258= sin(θ+450)θ+45=46.50θ=46.50−45=1.50

Conclusion:

1.500  the angle made by corner A of square box and floor.