A diver having a moment of inertia of $6.0 k g = m^{2}$ about an axis through its centre of mass rotates at an angular speed of 2 rad/s about this axis. If he folds his hands and feet to the decrease the moment of inertia to $5.0 k g - m^{2},$ what will be the new angular speed ?

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Solution

Given:
Initial Moment of inertia $I_{1} = 6 k g - m^{2}$
The initial angular velocity is $ω_{1} = 2 r a d / s$
Final moment of inertia $I_{2} = 5 k g - m^{2}$

Since external torque on the system is $= 0$
Therefore
$I_{1} ω_{1} = I_{2} ω_{2}$

$ω_{2} = \frac{I_{1} ω_{1}}{I_{2}}$

$\Rightarrow ω_{2} = \frac{6 \times 2}{5} = 2.4 r a d / s$

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A diver having a moment of inertia of 6.0 kg=m2 about an axis through its centre of mass rotates at an angular speed of 2 rad/s about this axis. If he folds his hands and feet to the decrease the moment of inertia to 5.0 kg−m2, what will be the new angular speed ?

Given:Initial Moment of inertia I1=6 kg−m2The initial angular velocity is ω1=2 rad/sFinal moment of inertia I2=5 kg−m2Since external torque on the system is =0Therefore I1ω1=I2ω2ω2=I1ω1I2⇒ω2=6×25=2.4 rad/s

A diver having a moment of inertia of $6.0 k g = m^{2}$ about an axis through its centre of mass rotates at an angular speed of 2 rad/s about this axis. If he folds his hands and feet to the decrease the moment of inertia to $5.0 k g - m^{2},$ what will be the new angular speed ?