Chapter 17.3, Problem 17.111P

A uniform slender rod of length L is dropped onto rigid supports at A and B. Because support B is slightly lower than support A, the rod strikes A with a velocity $\bar{v}$ before it strikes B. Assuming perfectly elastic impact at both A and B, determine the angular velocity of the rod and the velocity of its mass center immediately after the rod (a) strikes support A, (b) strikes support B, (c) again strikes support A.

  

To determine

(a)

Calculate the angular velocity and mass center velocity of rob when rod strikes support A.

Answer to Problem 17.111P

While rod strikes support A,

Angular velocity of rod $=\left(3v_1\right)/L\downarrow$

Mass center velocity of rod $=v_1/L\downarrow$

Explanation of Solution

For impact at A

$E=1and\left(v_A\right)=v_1\uparrow$

As per kinematics,

$V_2=L/2{\omega }_2-{\left(v_A\right)}_2=L/2{\omega }_2-v_1$

Taking moment at A,

$\begin{array}{l}m{V}_{1}L/2+0=m{v}_{2}L/2+I{\omega }_2\hfill \\ m{V}_{1}L/2=m\left[1/2\omega -v_1\right]L/2+\left[1/12m{L}^2\right]I{\omega }_2\hfill \end{array}$

Further solving, we get

${\omega }_2=3v_1/L\downarrow$

and mass center velocity of rod,

$\begin{array}{l}{V}_2=L/2{\omega }_2-v_1\hfill \\ =L/2\left(3v_1/L\right)-v_1\hfill \\ =3/2v_1-v_1\hfill \\ =v_1\left(3/2-1\right)\hfill \\ V_2=1/2v_1\hfill \end{array}$

Conclusion:

Rod having length is supported by both end at A and B support B is lower than support A. Hence, rod with velocity v₁strikes A before striking to B, assuming perfect elasticity between A and B.

Angular velocity of rod is $\left(3v_1\right)/L\downarrow$ and Mass centre velocity of rod is $v_1/L\downarrow$.

To determine

Calculate the angular velocity and mass center velocity of the rod when rod stricken supports B.

Answer to Problem 17.111P

When rod strikes support B,

The angular velocity of rod = $\left(3v_{1}0/L\right)\uparrow$

The mass center velocity of rod = $\left(v_1/2\right)\uparrow$

Explanation of Solution

$E=1and{\left(v_B\right)}_2=2v_1$

As per kinematics,

$\begin{array}{l}V={\left(v_B\right)}_2-L/2\omega \hfill \\ =2v_1-1/2\omega \hfill \end{array}$

Taking moment at B₁

$\begin{array}{l}-m{v}_{2}L/2+I\omega +0=m{v}_{3}L/3-I{\omega }_3\\ \begin{array}{l}\left(-mL/2x{v}_1/2\right)+(1/12m{L}^{2}x3v_1/L=m\left(2v_1-L/2{\omega }_3\right)L/2-\left(1/12m{L}^2\right){\omega }_3\hfill \\ -mL{v}_1/4+3v_{1}m{L}^2/12L=2mL{v}_1/2-m{\omega }_3{L}^2/4-m{L}^2{\omega }_3/12\hfill \end{array}\\ \begin{array}{l}-mL{v}_1/4+m{L}^2{v}_1/4=mL{v}_1-m{L}^2{\omega }_3/4-m{L}^2{\omega }_3/12\hfill \\ m{L}^2{\omega }_3/4+m{L}^2{\omega }_3/12=mL{v}_1\hfill \end{array}\\ \begin{array}{l}m{L}^2{\omega }_3/4\left(1+1/3\right)=mL{v}_1\hfill \\ {\omega }_3=3/4xmL{v}_{1}x4/m{L}^2\hfill \\ {\omega }_3=3v_1/L\hfill \end{array}\end{array}$

Now, mass center velocity of rod,

$v_3=2v_1-L/2\left(3v_1/L\right)=v_1/2$

Conclusion:

Rod having length is supported by both end at A and B support B is lower than support A. Hence rod with velocity v₁ strikes A before striking to B. assuming perfect elasticity between A and B.

The angular velocity of rod is $\left(3v_{1}0/L\right)\uparrow$ and the mass center velocity of rod is $\left(v_1/2\right)\uparrow$.

To determine

Calculate the angular velocity and mass center velocity of the rod when rod strikes A again?

Answer to Problem 17.111P

When rod strikes A again

The angular velocity of rod = $0$

The mass center velocity of rod = $v_1\uparrow$

Explanation of Solution

For impact A again,

$E=1and\left(v_A\right)=v_1$

As per kinematics,

$V_4={\left(v_A\right)}_4+L/2{\omega }_4=v_1+L/2{\omega }_4$

Taking moment about A,

$\begin{array}{l}\begin{array}{l}m{v}_{3}L/2+I{\omega }_3+0=m{v}_{4}L/4+I{\omega }_4\hfill \\ m\left(v_1/2\right)L/2+\left(1/12m{L}^2{\omega }_3\right)=m\left(v_1+L/2{\omega }_4\right)L/2+\left(1/12m{L}^2\right){\omega }_4\hfill \\ m{v}_{1}L/4+m{L}^2{\omega }_3/12=m{v}_{1}L/2+m{L}^2{\omega }_4/4+m{L}^2{\omega }_4/12\hfill \\ m{v}_{1}L/4+3m{L}^2\omega 3/12=m{v}_{1}L/2+m{L}^2{\omega }_4/4\left(1+1/3\right)\hfill \end{array}\\ \begin{array}{l}m{v}_{1}L/4+mvL/4-m{v}_{1}L/2=4/3\left(m{L}^2{\omega }_4/4\right)\hfill \\ 2\left(m{v}_{1}L/4\right)-m{v}_{1}L/2=4/3\left(m{L}^2{\omega }_4/4\right)\hfill \\ m{v}_{1}L/2\left(2/2-1\right)=4/3\left(m{L}^2{\omega }_4/4\right)\hfill \\ m{v}_{1}Lx3x4/2x4xm{L}^2={\omega }_4\hfill \\ 0={\omega }_4\hfill \end{array}\end{array}$

And mass center velocity of rod,

$\begin{array}{l}{v}_1=v_1+0\hfill \\ v_1=v_1\hfill \end{array}$

Conclusion:

Rod having length is supported by both end at A and B support B is lower than support A. Hence rod with velocity v₁ strikes A before striking to B assuming perfect elasticity between A and B.

The angular velocity of rod is $0$ and the mass center velocity of rod is $v_1\uparrow$.