Answer
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Hint: Angular velocity is also vector quantity equal to the angular displacement or vector quantity divided by the change in time is called angular displacement. By using this formula only we are calculating the change in its kinetic energy.
Complete step by step answer:
Here given the length of the angular momentum is $80\,kg.{m^2}\,{\sec ^{ - 1}}$ and the angular velocity of a body changes from $20\,rad\,{\sec ^{ - 1}}$ to $40\,rad\,{\sec ^{ - 1}}$. The length of the angular momentum is derived as,
$\Delta L = {L_2} - {L_1} = 80\,kg.{m^2}\,{\sec ^{ - 1}}$
Where as ${L_2} - {L_1} = I\left( {{\omega _2} - {\omega _1}} \right)$
From the above equations we are finding the change in kinetic energy.
Thus the equation of kinetic energy is,
$\Delta K.E = \dfrac{1}{2}I\left( {{\omega _2}^2 - {\omega _1}^2} \right)$
From the above equation we are found that the,
$\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$
From the above equation,
$\Delta K.E = \dfrac{1}{2}I\left( {{\omega _2} - {\omega _1}} \right)\left( {{\omega _2} + {\omega _1}} \right)$
By substituting all the given values in the change in kinetic energy is,
$\Delta K.E = \dfrac{1}{2}\left( {{L_2} - {L_1}} \right)\left( {{\omega _2} + {\omega _1}} \right) \\
\Rightarrow \Delta K.E = \dfrac{1}{2}\left( {80} \right)\left( {40 + 20} \right) \\
\Rightarrow \Delta K.E = \left( {40} \right)\left( {60} \right) \\
\therefore \Delta K.E = 2400\,J \\ $
From the angular momentum and angular velocities we have proved the change in its kinetic energy. For this we have taken the equation of angular displacement and kinetic energy formulas whereas kinetic energy is displaced by its energy of motion as the moving object or particle observes some energy in its motion, called kinetic energy.
Hence, the change in kinetic energy is $2400\,J$.
Note: From the given data we have angular velocity and angular momentum using this data and angular displacement formula and kinetic energy formula we have proved the given problem thus the kinetic energy is measured in joules. So the change in its kinetic energy is $2400$ joules.
Complete step by step answer:
Here given the length of the angular momentum is $80\,kg.{m^2}\,{\sec ^{ - 1}}$ and the angular velocity of a body changes from $20\,rad\,{\sec ^{ - 1}}$ to $40\,rad\,{\sec ^{ - 1}}$. The length of the angular momentum is derived as,
$\Delta L = {L_2} - {L_1} = 80\,kg.{m^2}\,{\sec ^{ - 1}}$
Where as ${L_2} - {L_1} = I\left( {{\omega _2} - {\omega _1}} \right)$
From the above equations we are finding the change in kinetic energy.
Thus the equation of kinetic energy is,
$\Delta K.E = \dfrac{1}{2}I\left( {{\omega _2}^2 - {\omega _1}^2} \right)$
From the above equation we are found that the,
$\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$
From the above equation,
$\Delta K.E = \dfrac{1}{2}I\left( {{\omega _2} - {\omega _1}} \right)\left( {{\omega _2} + {\omega _1}} \right)$
By substituting all the given values in the change in kinetic energy is,
$\Delta K.E = \dfrac{1}{2}\left( {{L_2} - {L_1}} \right)\left( {{\omega _2} + {\omega _1}} \right) \\
\Rightarrow \Delta K.E = \dfrac{1}{2}\left( {80} \right)\left( {40 + 20} \right) \\
\Rightarrow \Delta K.E = \left( {40} \right)\left( {60} \right) \\
\therefore \Delta K.E = 2400\,J \\ $
From the angular momentum and angular velocities we have proved the change in its kinetic energy. For this we have taken the equation of angular displacement and kinetic energy formulas whereas kinetic energy is displaced by its energy of motion as the moving object or particle observes some energy in its motion, called kinetic energy.
Hence, the change in kinetic energy is $2400\,J$.
Note: From the given data we have angular velocity and angular momentum using this data and angular displacement formula and kinetic energy formula we have proved the given problem thus the kinetic energy is measured in joules. So the change in its kinetic energy is $2400$ joules.
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