Question 258: Centre of Mass and Rotational Motion

Question

Two coaxial discs, having moments of inertia
I
1
and I
1
/2, are rotating with respective angular
velocities $$\omega $$
1
and
$$\omega $$
1
/2
, about their common axis.
They are brought in contact with each other and
thereafter they rotate with a common angular
velocity. If E
f
and E
i
are the final and initial total
energies, then  (E
f
- E
i
) is:

Accepted Answer

$${E_i} = {1 \over 2}{I_I} 	imes \omega _1^2 + {1 \over 2}{I \over 2} 	imes {{\omega _1^2} \over 4}$$
$$ = {{{I_1}\omega _1^2} \over 2}\left( {{9 \over 8}} ight) = {9 \over {16}}{I_1}\omega _1^2$$
$${I_1}{\omega _1} + {{{I_1}{\omega _1}} \over 4} = {{3{I_1}} \over 2}\omega ;{5 \over 4}{I_1}{\omega _1} = {{3{I_1}} \over 2}\omega $$
$$\omega  = {5 \over 6}{\omega _1};{E_f} = {1 \over 2} 	imes {{3{I_1}} \over 2} 	imes {{25} \over {36}}\omega _1^2$$
$$ = {{25} \over {48}}{I_1}\omega _1^2$$
$$ \Rightarrow {E_f} - {E_i} = {I_1}\omega _1^2{{25} \over {49}} - {{ - 2} \over {48}}{I_2}\omega _1^2$$
$$ = {{25} \over {48}}{I_1}\omega _1^2$$
$$ \Rightarrow {E_f} - {E_i} = {I_1}\omega _1^2\left( {{{25} \over {48}} - {9 \over {16}}} ight) = {{ - 2} \over {48}}{I_1}\omega _1^2$$
$$ = {{ - {I_1}\omega _1^2} \over {24}}$$

Answer

$${{{I_1}\omega _1^2} \over {24}}$$

Answer

$${{{I_1}\omega _1^2} \over {12}}$$

Answer

$${3 \over 8}{I_1}\omega _1^2$$

Answer

$${{{I_1}\omega _1^2} \over {6}}$$

Centre of Mass and Rotational Motion

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