Chapter 17.3, Problem 17.118P

A uniformly loaded square crate is released from rest with its comer D directly above A; it rotates about A until its comer B strikes the floor, and then rotates about B. The floor is sufficiently rough to prevent slipping and the impact at B is perfectly plastic. Denoting by ${\omega }_0$ the angular velocity of the crate immediately before B strikes the floor, determine (a) the angular velocity of the crate immediately after B strikes the floor, (b) the fraction of the kinetic energy ofthe crate lost during the impact, (c) the angle $\theta$ through which the crate will rotate after B strikes the floor.

  

To determine

(a)

The angular velocity of the crate immediately after B strikes the floor.

Answer to Problem 17.118P

The angular velocity of the square box after striking B on the floor will be $\frac{1}{4}$ th of angular velocity before B strikes on the floor.

$\omega =1/4{\omega }_0$

Explanation of Solution

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 = $I=1/2m\left(c^2+c^2\right)=1/6m{c}^2$

We can say,

$v_0=r_{G/A}x{\omega }_0=1/2{\omega }_0{}_{}v=r_{G/B}x\omega =1/2\omega$

Taking moment at A,

$\begin{array}{l}I{\omega }_0+0=I\omega +r_{G/B}xmv1/6m{c}^2{\omega }_0\\ =1/6m{c}^2\omega +(1/2)m(1/2)1/6m{c}^2{\omega }_0\\ =1/6m{c}^2\omega +[/4]\\ \begin{array}{l}1/6m{c}^2{\omega }_0=1/6m{c}^2\omega +2c^{2}m\omega /4\hfill \\ mc2{\omega }_0/6=m{c}^2\omega /6+2c^{2}m\omega /4\hfill \\ mc2{\omega }_0/6=m{c}^2\omega /2\left(1/3+2/2\right)\hfill \\ mc2{\omega }_0/6=4/1\left(m{c}^2\omega /2\right)\hfill \\ 3/4x2/6xm{c}^2{\omega }_0/m{c}^2=\omega \hfill \\ 1/4{\omega }_0=\omega \hfill \end{array}\end{array}$

Conclusion:

The angular velocity of the square box after striking B on the floor will be $\frac{1}{4}$ of the angular velocity before B strikes on the floor.

To determine

(b)

The fraction of kinetic energy lost during the impact.

Answer to Problem 17.118P

The fractions of energy lost during impact conditions are.

Explanation of Solution

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,

$\begin{array}{l}{E}_1=1/2I{\omega }_0{}^2+1/2m{v}_0{}^2\\ =1/2\left(1/6m{c}^2\right){\omega }_0{}^2+1/2m{(}^1\\ \begin{array}{l}=m{c}^2{\omega }_0{}^2/12+m/2\left(2/4c^2{\omega }_0{}^2\right)\hfill \\ =m{c}^2{\omega }_0{}^2/12+m{c}^2{\omega }_0{}^2/4\hfill \\ =m{c}^2{\omega }_0{}^2\left(1/12+1/4\right)\hfill \\ E_1=1/3m{c}^2{\omega }_0{}^2\hfill \end{array}\end{array}$

Similarly kinetic energy after impact,

$\begin{array}{l}{E}_2=1/2I{\omega }^2+1/2m{v}^2\\ =1/2\left(1/6m{c}^2\right){\omega }^2+1/2m{(}^1\\ \begin{array}{l}=m{c}^2{\omega }^2/12+m{c}^2{\omega }^2/4\hfill \\ =1/3m{c}^2{\omega }^2\hfill \\ =1/3m{c}^2{\left(1/4{\omega }_0\right)}^2\hfill \\ E_2=1/48m{c}^2{\omega }_0{}^2\hfill \end{array}\end{array}$

Combining E₁ and E₂ to find fraction of energy cost,

$\begin{array}{l}{E}_1-E_2/E_1=1/3m{c}^2{\omega }_0–1/48m{c}^2{\omega }_0/1/3m{c}^2{\omega }_0\hfill \\ =\left(1/3-1/48\right)m{c}^2{\omega }_0{}^2/\left(1/3\right)m{c}^2{\omega }_0{}^2\hfill \\ =1/3-1/48/1/3\hfill \\ =(1/3/\left(1/3\right)-\left(1/48\right)/\left(1/3\right)\hfill \\ =3/3-3/48\hfill \\ =1-1/16\hfill \\ 15/16\hfill \end{array}$

Conclusion:

The fraction of energy lost during impact conditions is $\frac{15}{16}$.

To determine

(c)

The angle through which the crater will rotate after B strikes the floor.

Answer to Problem 17.118P

Angle made by corner A of square box and floor will be 1.50⁰

Explanation of Solution

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,

$T_0+V_0=T_1+V_1$

And that of after impact,

$\begin{array}{l}{T}_3+V_3=T_2+V_2\\ T_0=0,\\ T_3=0,\\ T_2=1/16T{v}_0\\ =mg(1/2\sqrt{2}c)v_1=v_2\\ =mg\left(1/2\sqrt{2}c\right)v_3\\ =mg{h}_3\end{array}$

Before impact,

$\begin{array}{l}\begin{array}{l}{T}_0+V_0=T_1+V_1\hfill \\ T_1=V_0-V_1\hfill \end{array}\\ =mg(1/2\sqrt{2}c)\\ =mg(1/2\sqrt{2}c)-mg\left(1/2c\right)\\ =mg\sqrt{2}c/2-mgc/2T_1\\ =mgc/2(\sqrt{2}-1)\end{array}$

After impact,

$\begin{array}{l}{T}_3+V_3=T_2+V_2\hfill \\ T_2=V_3-V_2\hfill \\ =mg{h}_3-mgc/2\hfill \\ T_2=mg\left(h_3-c/2\right)\hfill \end{array}$

As,

$\begin{array}{l}{T}_2=1/16T_1\\ 1/16mgc/2(\sqrt{2}-1)=mg\left(h_3-c/2\right)\\ \sqrt{2}-1/32c=h_3-c/2\\ h_3=((\sqrt{2}-1)c/32+c/2\\ =((\sqrt{2}-1)c+16c/32\\ h_3=((\sqrt{2}-1)+16)c/32\end{array}$

But

Therefore, from geometry,

$h_3=1/2\sqrt{2}csin\left(\theta +{45}^0\right)$

Equating both equations of h₃

$\begin{array}{l}[(\sqrt{2}-1)+16/32]c=\sqrt{2}c/2sin\left(\theta +{45}^0\right)\\ [(\sqrt{2}-1)+16/32]c\times 2/\sqrt{2}c=sin\left(\theta +{45}^0\right)\\ \sqrt{2}-1/16+1=sin\left(\theta +{45}^0\right)\\ \begin{array}{l}1.0258=sin\left(\theta +{45}^0\right)\hfill \\ \theta +45=46.50\hfill \\ \theta =46.50-45={1.5}^0\hfill \end{array}\end{array}$

Conclusion:

${1.50}^{0 }$ the angle made by corner A of square box and floor.