A diver having a moment of inertia of 6.0 ${kg-m}^{2}$ 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 decrease the moment of inertia to 5.0 ${kg-m}^{2}$ , what will be the new angular speed ?

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Solution

$I = 6 {kg-m}^{2}, ω = 2 rad/s, I_{2} = 5 {kg-m}^{2}$

$Since, external torque = 0$

$Therefore, I_{1} ω_{1} = I_{2} ω_{2}$

$\Rightarrow ω_{2} = \frac{(6 \times 2)}{5}$

$= \frac{12}{5} = 2.4 rad/s$

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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 decrease the moment of inertia to 5.0 kg-m2, what will be the new angular speed ?

I=6kg-m2,ω=2rad/s,I2=5kg-m2 Since, external torque=0 Therefore, I1ω1=I2ω2 ⇒ ω2=(6×2)5 =125=2.4rad/s

A diver having a moment of inertia of 6.0 ${kg-m}^{2}$ 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 decrease the moment of inertia to 5.0 ${kg-m}^{2}$ , what will be the new angular speed ?

$I = 6 {kg-m}^{2}, ω = 2 rad/s, I_{2} = 5 {kg-m}^{2}$

$Since, external torque = 0$

$Therefore, I_{1} ω_{1} = I_{2} ω_{2}$

$\Rightarrow ω_{2} = \frac{(6 \times 2)}{5}$

$= \frac{12}{5} = 2.4 rad/s$