Question

Two $\mathbf{2}\mathbf{.00}\mathbf{}\mathbf{kg}$balls are attached to the ends of a thin rod of length role="math" localid="1661007264498" $\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$and negligible mass. The rod is free to rotate in a vertical plane without friction about a horizontal axis through its center. With the rod initially horizontal (In Figure), $\mathbf{}\mathbf{50.0}\mathbf{}\mathbf{g}$wad of wet putty drops onto one of the balls, hitting it with a speed of$\mathbf{3.00}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$ and then sticking to it.

(a) What is the angular speed of the system just after the putty wad hits?

(b) What is the ratio of the kinetic energy of the system after the collision to that of the putty wad just before?

(c) Through what angle will the system rotate before it momentarily stops?

Short answer

1. Angular speed of the system just after the putty wad hitsis$\mathrm{\omega }=0.148\text{rad}/\text{s}$.
2. Ratio of kinetic energy of the system after the collision to just before is$\frac{{\text{K}}_{\text{f}}}{{\text{K}}_{\text{i}}}=0.0123$.
3. Angle through which system rotates before it stops is $\text{θ}={181}^0$.

Step-by-step solution

Step 1: Given

1. Mass of the ball$\left(\mathrm{M}\right)=2\text{\hspace{0.17em}\hspace{0.17em}kg}$
2. Mass of putty$\left(\mathrm{m}\right)=0.05\text{kg}$
3. Speed$\left(\mathrm{v}\right)=3\text{m}/\text{s}$
4. $\mathrm{d}=0.5\mathrm{m},\mathrm{r}=0.25\mathrm{m}$

Determining the concept

Apply law of conservation of angular momentum to find the angular speed. Then, calculate angle by using law of conservation of energy. According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula are as follow:

1. Initial angular momentum = Final angular momentum
2. $\mathbf{L}\mathbf{=}\mathbf{}\mathbf{mvr}$

Where,$\mathbf{L}$is angular momentum, $\mathbf{}\mathbf{m}$is mass,$\mathbf{v}$ is velocity and $\mathbf{r}$ is radius.

 Determining the angular speed of the system just after the putty wad hits

(a)

According to law of conservation of momentum :

Initial angular momentum = Final angular momentum

$\begin{array}{l}{\mathrm{L}}_{\mathrm{i}}={\mathrm{L}}_{\mathrm{f}}\\ {\mathrm{L}}_{\mathrm{i}}=\mathrm{mvr}=0.05\times 3\times 0.25\\ {\mathrm{L}}_{\mathrm{i}}=0.03750\text{kg}.{\text{m}}^2/\text{s}\\ {\mathrm{L}}_{\mathrm{f}}={\mathrm{I\omega }}_{\mathrm{f}}\end{array}$

$\begin{array}{l}\mathrm{I}=2{\mathrm{Mr}}^2+{\mathrm{mr}}^2\\ \mathrm{I}=(2\times 2\times {0.25}^2)+(0.05\times {0.25}^2)\\ \mathrm{I}=0.253125\end{array}$

Hence,

$\begin{array}{c}0.03750=0.253125\times {\mathrm{\omega }}_{\mathrm{f}}\\ {\mathrm{\omega }}_{\mathrm{f}}=0.148\text{rad}/\text{s}\end{array}$

Hence, angular speed of the system just after the putty wad hits is $\mathrm{\omega }=0.148\text{rad}/\text{s}$.

Determining the ratio of kinetic energy of the system after/before the collision to the putty wad

(b)

$\text{Initial}\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2$

$\text{Final}\mathrm{KE}=\frac{1}{2}{\mathrm{I\omega }}^2$

$\frac{{\mathrm{KE}}_{\mathrm{f}}}{{\mathrm{KE}}_{\mathrm{i}}}=\frac{{\mathrm{mv}}^2}{{\mathrm{I\omega }}^2}$

$\frac{{\mathrm{KE}}_{\mathrm{f}}}{{\mathrm{KE}}_{\mathrm{i}}}=\frac{0.253125\times 0{.148}^2}{0.05\times 3^2}=0.0123$

Hence, ratio of kinetic energy of the system after the collision to just before is$\frac{{\text{K}}_{\text{f}}}{{\text{K}}_{\text{i}}}=0.0123$ .

Determining the angle through which system rotates before it stops

(c)

As the rod rotates, according to law of conservation of energy sum of kinetic energy and potential energy remains constant. If one ball is lowered by distanceh, other one would be raised by the same distance. So, the sum of the energies of the balls would never change. If the putty wad is considered, to reach the bottom, it would travel through an angle$90°$with horizontal. During this, it would lose itsPE and gainKE. Then, it would go up, reversing the process and gainingPE and losingKE, until it stops for a moment. Let’s assume that it swings up through the angle$\mathrm{\theta }$. If the bottom point is taken as origin, then its initial potential energy is,

${\mathrm{PE}}_{\mathrm{i}}=\frac{\mathrm{mgd}}{2}$

If it swings by an angle $\mathrm{\theta }$,

So,

$\mathrm{h}=\frac{\mathrm{d}}{2}-\frac{\mathrm{d}}{2}\mathrm{cos\theta }$

$\mathrm{h}=\frac{\mathrm{d}}{2}(1-\mathrm{cos\theta })$

Using law of conservation of energy,

$\begin{array}{c}{\mathrm{KE}}_{\mathrm{i}}+{\mathrm{PE}}_{\mathrm{i}}={\mathrm{KE}}_{\mathrm{f}}+{\mathrm{PE}}_{\mathrm{f}}\\ \left(\frac{1}{2}\right){\mathrm{I\omega }}^2+\mathrm{mg}\left(\frac{\mathrm{d}}{2}\right)=\mathrm{mg}\left(\frac{\mathrm{d}}{2}\right)(1-\mathrm{cos\theta })\\ \left(\frac{1}{2}\right)0.253125\times 0{.148}^2+0.05\times 9.8\times \left(\frac{0.5}{2}\right)=0.05\times 9.8\times \left(\frac{0.5}{2}\right)\times (1-\cos \left(\mathrm{\theta }\right))\\ 2.772\times {10}^{-3}+1.225=1.225\times (1-\cos \left(\mathrm{\theta }\right))\end{array}$

Hence,

$\begin{array}{c}1-\mathrm{cos\theta }=1.00226\\ \mathrm{cos\theta }=-0.00226\\ \mathrm{\theta }={\cos }^{-1}(-0.00226)={91}^0\end{array}$

$\text{Totalangle}={90}^0+{91}^0={181}^0$

Hence, angle through which system rotates before it stops is$\text{θ}={181}^0$ .

Therefore, using the law of conservation of angular momentum and conservation of energy together, angular speed, ratio of kinetic energy and angle of rotation before stopping can be found.