91Ó°ÊÓ

1. Mass of the ball$\left(\mathrm{M}\right)=2\text{ â¶Ä‰k²µ}$
2. Mass of putty$\left(\mathrm{m}\right)=0.05\text{kg}$
3. Speed$\left(\mathrm{v}\right)=3\text{m}/\text{s}$
4. $\mathrm{d}=0.5\mathrm{m},\mathrm{r}=0.25\mathrm{m}$

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.

(a)

According to law of conservation of momentum :

Initial angular momentum = Final angular momentum

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

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

Hence,

$\begin{array}{c}0.03750=0.253125×{\mathrm{Ó\neg }}_{\mathrm{f}}\\ {\mathrm{Ó\neg }}_{\mathrm{f}}=0.148\text{rad}/\text{s}\end{array}$

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

(b)

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

$\text{Final}\mathrm{KE}=\frac{1}{2}{\mathrm{IÓ\neg }}^2$

$\frac{{\mathrm{KE}}_{\mathrm{f}}}{{\mathrm{KE}}_{\mathrm{i}}}=\frac{{\mathrm{mv}}^2}{{\mathrm{IÓ\neg }}^2}$

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

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

(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$\mathrm{θ}$. 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 $\mathrm{θ}$,

So,

$\mathrm{h}=\frac{\mathrm{d}}{2}−\frac{\mathrm{d}}{2}\mathrm{³¦´Ç²õθ}$

$\mathrm{h}=\frac{\mathrm{d}}{2}(1−\mathrm{³¦´Ç²õθ})$

Using law of conservation of energy,

$\begin{array}{c}{\mathrm{KE}}_{\mathrm{i}}+{\mathrm{PE}}_{\mathrm{i}}={\mathrm{KE}}_{\mathrm{f}}+{\mathrm{PE}}_{\mathrm{f}}\\ \left(\frac{1}{2}\right){\mathrm{IÓ\neg }}^2+\mathrm{mg}\left(\frac{\mathrm{d}}{2}\right)=\mathrm{mg}\left(\frac{\mathrm{d}}{2}\right)(1−\mathrm{³¦´Ç²õθ})\\ \left(\frac{1}{2}\right)0.253125×0{.148}^2+0.05×9.8×\left(\frac{0.5}{2}\right)=0.05×9.8×\left(\frac{0.5}{2}\right)×(1−\cos \left(\mathrm{θ}\right))\\ 2.772×{10}^{−3}+1.225=1.225×(1−\cos \left(\mathrm{θ}\right))\end{array}$

Hence,

$\begin{array}{c}1−\mathrm{³¦´Ç²õθ}=1.00226\\ \mathrm{³¦´Ç²õθ}=−0.00226\\ \mathrm{θ}={\cos }^{−1}(−0.00226)={91}^0\end{array}$

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

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.