JEE MAIN - Physics (2020 - 5th September Morning Slot - No. 12)
A wheel is rotating freely with an angular speed
$$\omega $$ on a shaft. The moment of inertia of the
wheel is I and the moment of inertia of the
shaft is negligible. Another wheel of moment of
inertia 3I initially at rest is suddenly coupled to
the same shaft. The resultant fractional loss in
the kinetic energy of the system is :
0
$${5 \over 6}$$
$${1 \over 4}$$
$${3 \over 4}$$
توضیح
Applying Angular Momentum conservation
$$I\omega = (I + 3I)\omega '$$
$$ \Rightarrow $$ $$\omega ' = {{I\omega } \over {4I}} = {\omega \over 4}$$
$${k_i} = {1 \over 2}I{\omega ^2}$$
$${k_f} = {1 \over 2}(4I){(\omega ')^2}$$
$$ = 2I{\left( {{\omega \over 4}} \right)^2} = {1 \over 8}I{\omega ^2}$$
Fractional loss $$ = {{{K_i} - {K_f}} \over {{K_i}}} = {{{1 \over 2}I{\omega ^2} - {1 \over 8}I{\omega ^2}} \over {{1 \over 2}I{\omega ^2}}}$$
= $${{{3 \over 8}I{\omega ^2}} \over {{1 \over 2}I{\omega ^2}}} = {3 \over 4}$$
$$I\omega = (I + 3I)\omega '$$
$$ \Rightarrow $$ $$\omega ' = {{I\omega } \over {4I}} = {\omega \over 4}$$
$${k_i} = {1 \over 2}I{\omega ^2}$$
$${k_f} = {1 \over 2}(4I){(\omega ')^2}$$
$$ = 2I{\left( {{\omega \over 4}} \right)^2} = {1 \over 8}I{\omega ^2}$$
Fractional loss $$ = {{{K_i} - {K_f}} \over {{K_i}}} = {{{1 \over 2}I{\omega ^2} - {1 \over 8}I{\omega ^2}} \over {{1 \over 2}I{\omega ^2}}}$$
= $${{{3 \over 8}I{\omega ^2}} \over {{1 \over 2}I{\omega ^2}}} = {3 \over 4}$$