Chapter 10, Problem 68P

(a)

To determine

The mass when a particle is stick to rod.

Answer to Problem 68P

  $1.2kg$

Explanation of Solution

Given:

Mass of rod, M = $2kg$

  $\begin{array}{l}{L}_1=1.2m\\ L_2=0.80m\\ {\theta }_{\max }={37}^o\end{array}$

Formula used:

Conservation of mechanical energy:

  $K_f-K_i+U_f-U_i=0$

  $K_f$ : Final kinetic energy

  $K_i$ : Initial kinetic energy

  $U_f$ : Final potential energy

  $U_i$: Initial potential energy

Rotational kinetic energy:

  $K_r=\frac{1}{2}I{\omega }^2$

Where, Iis the moment of inertia and $\omega$ is the angular velocity.

Calculation:

Consider $L_1$ to be the distance of zero of gravitational potential energy below the pivot.

  $\begin{array}{l}{K}_f-K_i+U_f-U_i=0\\ \Rightarrow K_f+U_f-U_i=0\left(\because K_i=0\right)\end{array}$

Substitute $K_f,U_f\&U_i$

  $\frac{1}{2}\left(\frac{1}{3}M{L}_1^2\right){\omega }^2+Mg\frac{L_1}{2}-Mg{L}_1=0$

  $\Rightarrow \omega =\sqrt{\frac{3g}{L_1}}$

Consider the angular speed of the system after impact to be $\omega \prime$

As angular momentum is conserved, so $\Delta L=L_f-L_i=0$

  $\Rightarrow \frac{1}{2}\left(\frac{1}{3}M{L}_1^2+m{L}_2^2\right)\omega \text{'}-\left(\frac{1}{3}M{L}_1^2\right)\omega =0$

So $\begin{array}{l}\omega \text{'}=\frac{\frac{1}{3}M{L}_1^2}{\left(\frac{1}{3}M{L}_1^2+m{L}_2^2\right)}\omega \\ =\frac{\frac{1}{3}M{L}_1^2}{\left(\frac{1}{3}M{L}_1^2+m{L}_2^2\right)}\times \sqrt{\frac{3g}{L_1}}\end{array}$

Substitute numerical values and simplify to obtain:

  $\begin{array}{l}\omega \text{'}=\frac{\frac{1}{2}(2kg){\left(1.2\right)}^2}{\frac{1}{3}(2kg){\left(1.2\right)}^2+m{\left(0.80m\right)}^2}\times \sqrt{\frac{3\times \left(9.81m/s^2\right)}{1.2m}}\\ \omega \text{'}=\frac{4.75kg}{0.96kg+0.64m}\end{array}$

The mechanical energy is conserved. The rotational kinetic energy of the system just after their collision can be related to their potential energy.

  $\begin{array}{l}{K}_f-K_i+U_f-U_i=0\\ \Rightarrow -K_i+U_f-U_i=0\left(\because K_f=0\right)\end{array}$

Substitute $K_i,U_f\&U_i$

  $-\frac{1}{2}I\omega {\text{'}}^{2}Mg\left(\frac{1}{2}{L}_1\right)\left(1-\cos {\theta }_{\max }\right)+mg{L}_2\left(1-\cos {\theta }_{\max }\right)=0$

  $I=\frac{1}{3}M{L}_1^2+m{L}_2^2$

Substitute ${\theta }_{\max },I\&\omega \text{'}$ :

  $\frac{\frac{\text{1}}{\text{2}}{\left(\text{4}\text{.75kg/s}\right)}^{\text{2}}}{\text{0}\text{.96 kg + 0}\text{.64 m}}=0.2g\left(M{L}_1+m{L}_2\right)$

Substituting the values of $M,L_1\text{and}{L}_2$ :

  $\frac{\frac{1}{2}{\left(\text{4}\text{.75kg/s}\right)}^2}{\text{0}\text{.96 kg+0}\text{.64 m}}=\text{0}\text{.2g}\left(\text{2}\text{.4 kg}\cdot \text{m+}m\left(\text{0}\text{.8m}\right)\right)$

  $\begin{array}{l}\Rightarrow m=1.18\text{kg}\\ \Rightarrow m𑨀1.2\text{kg}\end{array}$

Conclusion:

The mass when a particle is stick to rod is $1.2kg$ .

(b)

To determine

The energy dissipated in the inelastic collision.

Answer to Problem 68P

7.5J

Explanation of Solution

Given:

Mass of rod, M = $2kg$

  $\begin{array}{l}{L}_1=1.2m\\ L_2=0.80m\\ {\theta }_{\max }={37}^o\end{array}$

Formula used:

From the previous part:

  $\begin{array}{l}{U}_i=Mg\frac{L_1}{2}\\ U_f=\left(1-\cos {\theta }_{\max }\right)g\left(M\frac{L_1}{2}+m{L}_2\right)\end{array}$

Calculation:

The energy dissipated in the inelastic collision is: $\Delta E=U_i-U_f$

  $\begin{array}{l}{U}_i=Mg\frac{L_1}{2}\\ U_f=\left(1-\cos {\theta }_{\max }\right)g\left(M\frac{L_1}{2}+m{L}_2\right)\end{array}$

  $\Delta E=Mg\frac{L_1}{2}-\left[\left(1-\cos {\theta }_{\max }\right)g\left(M\frac{L_1}{2}+m{L}_2\right)\right]$

Substitute all the values and solve:

  $\begin{array}{l}\Delta E=\frac{\left(2kg\right)\left(9.81m/s^2\right)\left(1.2m\right)}{2}-\left[\left(1-\cos {37}^o\right)\left(9.81m/s^2\right)\left(\frac{\left(2kg\right)\left(1.2m\right)}{2}+\left(1.18kg\right)\left(0.80m\right)\right)\right]\\ \Delta E=7.5J\end{array}$

Conclusion:

The energy dissipated in the inelastic collision is 7.5J.