91Ó°ÊÓ

1. If the rotational inertia of first wheel is then the rotational inertia of second wheel is ${\text{I}}_2=2{\text{I}}_1$.
2. The initial angular speed of the first wheel is,${\text{Ó¬}}_1=800\text{rev}/\text{min}$
3. The initial angular speed of the second disk is,${\text{Ó¬}}_2=0\text{rev}/\text{min}$

Using the law of conservation of angular momentum, find the angular speed of the system after coupling. Using the formula for K.E, find the initial and final K.E of the system from which, find the fractionof the initial K.E lost after coupling of the wheels.According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula is as follow:

The law of conservation of angular momentum,

${\mathbf{L}}_{\mathbf{i}}\mathbf{}\mathbf{=}{\mathbf{L}}_{\mathbf{f}}$

Where, L_i, L_f are initial and finalmomentums.

(a)

The law of conservation of angular momentum gives,

${\mathrm{L}}_{\mathrm{i}}={\mathrm{L}}_{\mathrm{f}}$

Angular momentum of the system before coupling = Angular momentum of the system after coupling

$\begin{array}{c}{\mathrm{I}}_1{\mathrm{Ó\neg }}_1+{\mathrm{I}}_2{\mathrm{Ó\neg }}_2=({\mathrm{I}}_1+{\mathrm{I}}_2)\mathrm{Ó\neg }\\ \mathrm{Ó\neg }=\frac{{\mathrm{I}}_1{\mathrm{Ó\neg }}_1+{\mathrm{I}}_2{\mathrm{Ó\neg }}_2}{{\mathrm{I}}_1+{\mathrm{I}}_2}\\ \mathrm{Ó\neg }=\frac{{\mathrm{I}}_1{\mathrm{Ó\neg }}_1+2{\mathrm{I}}_1\left(0\right)}{{\mathrm{I}}_1+2{\mathrm{I}}_1}\\ \mathrm{Ó\neg }=\left(\frac{{\mathrm{I}}_1}{{\mathrm{I}}_1+2{\mathrm{I}}_1}\right){\mathrm{Ó\neg }}_1\end{array}$

$\begin{array}{l}\mathrm{Ó\neg }=\frac{1}{3}\left(800\right)\\ \mathrm{Ó\neg }=267\text{rev/min}\end{array}$

Hence, the angular speed of the system of the coupled wheels if the second disk is spinning clockwise is $267\text{rev}/\text{min}$.

(b)

The initial kinetic energy of the system is,

$\begin{array}{l}\mathrm{K}.{\mathrm{E}}_{\mathrm{i}}=\frac{1}{2}{\mathrm{I}}_1{\mathrm{Ó\neg }}_1^2+\frac{1}{2}{\mathrm{I}}_2{\mathrm{Ó\neg }}_2^2\\ \mathrm{K}.{\mathrm{E}}_{\mathrm{i}}=\frac{1}{2}{\mathrm{I}}_1{\mathrm{Ó\neg }}_1^2+0\\ \mathrm{K}.{\mathrm{E}}_{\mathrm{i}}=\frac{1}{2}{\mathrm{I}}_1{\mathrm{Ó\neg }}_1^2\end{array}$

The final K.E of the system is,

role="math" localid="1660986935404" $\begin{array}{l}\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}=\frac{1}{2}({\mathrm{I}}_1+{\mathrm{I}}_2){\mathrm{Ó\neg }}^2\\ \mathrm{K}.{\mathrm{E}}_{\mathrm{f}}=\frac{1}{2}({\mathrm{I}}_1+2{\mathrm{I}}_1){\left[\left(\frac{{\mathrm{I}}_1}{{\mathrm{I}}_1+2{\mathrm{I}}_1}\right){\mathrm{Ó\neg }}_1\right]}^2\\ \mathrm{K}.{\mathrm{E}}_{\mathrm{f}}=\frac{1}{2}\left(3{\mathrm{I}}_1\right){\left(\frac{{\mathrm{I}}_1}{3{\mathrm{I}}_1}\right)}^2{\mathrm{Ó\neg }}_1^2\end{array}$

$\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}=\frac{1}{6}{\mathrm{I}}_1{\mathrm{Ó\neg }}_1^2$

So, the fraction of the energy lost after coupling is,

$\frac{\mathrm{Δ°­}.\mathrm{E}}{\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}}=\frac{(\mathrm{K}.{\mathrm{E}}_{\mathrm{i}}−\mathrm{K}.{\mathrm{E}}_{\mathrm{f}})}{\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}}$

$\frac{\mathrm{Δ°­}.\mathrm{E}}{\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}}=1−\frac{\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}}{\mathrm{K}.{\mathrm{E}}_{\mathrm{i}}}$

$\frac{\mathrm{Δ°­}.\mathrm{E}}{\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}}=1−\frac{\frac{1}{6}{\mathrm{I}}_1{\mathrm{Ó\neg }}_1^2}{\frac{1}{2}{\mathrm{I}}_1{\mathrm{Ó\neg }}_1^2}$

$\frac{\mathrm{Δ°­}.\mathrm{E}}{\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}}=\frac{2}{3}=0.667$

Hence, the fraction of the initial kinetic energy lost after coupling of the wheels is $0.667$.

Using conservation law of momentum, we can find the final angular speed of the system after coupling, if we know the initial angular speed and M.I of each object of the system. Using the formula for K.E, we can find the fraction of the initial K.E lost after coupling of the objects.