Question 174: Centre of Mass and Rotational Motion

Question

Two discs have moments of inertia I1 and I2 about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, $$\omega$$1 and $$\omega$$2 respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by :

Accepted Answer

From conservation of angular momentum we get
$${I_1}{\omega _1} + {I_2}{\omega _2} = ({I_1} + {I_2})\omega $$
$$\omega  = {{{I_1}{\omega _1} + {I_2}{\omega _2}} \over {{I_1} + {I_2}}}$$
$${k_i} = {1 \over 2}{I_1}\omega _1^2 + {1 \over 2}{I_2}\omega _2^2$$
$${k_f} = {1 \over 2}({I_1} + {I_2}){\omega ^2}$$
$${k_i} - {k_f} = {1 \over 2}\left[ {{I_1}\omega _1^2 + {I_2}\omega _2^2 - {{{{({I_1}{\omega _1} + {I_2}{\omega _2})}^2}} \over {{I_1} + {I_2}}}} \right]$$
Solving above we get
$${k_i} - {k_f} = {1 \over 2}\left( {{{{I_1}{I_2}} \over {{I_1} + {I_2}}}} \right){({\omega _1} - {\omega _2})^2}$$

Answer

$${{{I_1}{I_2}} \over {({I_1} + {I_2})}}{({\omega _1} - {\omega _2})^2}$$

Answer

$${{{{({I_1} - {I_2})}^2}{\omega _1}{\omega _2}} \over {2({I_1} + {I_2})}}$$

Answer

$${{{I_1}{I_2}} \over {2({I_1} + {I_2})}}{({\omega _1} - {\omega _2})^2}$$

Answer

$${{{{({\omega _1} - {\omega _2})}^2}} \over {2({I_1} + {I_2})}}$$

Centre of Mass and Rotational Motion

Choose the correct answer: