Question

A wheel is rotating freely at angular speed $\mathbf{800}\mathbf{}\mathbf{rev}\mathbf{/}\mathbf{min}$on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with twice the rotational inertia of the first, is suddenly coupled to the same shaft.

(a) What is the angular speed of the resultant combination of the shaft and two wheels?

(b) What fraction of the original rotational kinetic energy is lost?

Short answer

1. The angular speed of the system of the coupled wheels if the second disk is spinning clockwise is $267\text{rev}/\text{min}$.
2. The fraction of the initial kinetic energy lost after coupling of the wheels is $0.667$.

Step-by-step solution

Step 1: Given

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,${\text{ω}}_1=800\text{rev}/\text{min}$
3. The initial angular speed of the second disk is,${\text{ω}}_2=0\text{rev}/\text{min}$

Determining the concept

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.

Determining the angular speed of the system of the coupled wheels if the second disk is spinning clockwise

(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

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

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

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

 Determining the fraction of the initial kinetic energy lost after coupling of the wheels

(b)

The initial kinetic energy of the system is,

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

The final K.E of the system is,

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

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

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

$\frac{\mathrm{\Delta K}.\mathrm{E}}{\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}}=\frac{(\mathrm{K}.{\mathrm{E}}_{\mathrm{i}}-\mathrm{K}.{\mathrm{E}}_{\mathrm{f}})}{\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}}$

$\frac{\mathrm{\Delta K}.\mathrm{E}}{\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}}=1-\frac{\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}}{\mathrm{K}.{\mathrm{E}}_{\mathrm{i}}}$

$\frac{\mathrm{\Delta K}.\mathrm{E}}{\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}}=1-\frac{\frac{1}{6}{\mathrm{I}}_1{\mathrm{\omega }}_1^2}{\frac{1}{2}{\mathrm{I}}_1{\mathrm{\omega }}_1^2}$

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

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.