Answer:
The angular velocity just before landing on the trampoline bed is 5 [tex]\frac{rad}{s}[/tex]
Explanation:
Given:
Angular speed of gymnast [tex]\omega _{1} = 20\frac{rad}{s}[/tex]
From the formula of radius of gyration,
[tex]R = \sqrt{\frac{I}{A} }[/tex]
Where [tex]I =[/tex] moment of inertia
[tex]I = AR^{2}[/tex]
According to the conservation of angular momentum,
[tex]I_{1} \omega _{1} = I_{2} \omega _{2}[/tex]
[tex]AR_{1} ^{2} \omega _{1} = A R_{2} ^{2} \omega _{2}[/tex]
Here radius of gyration is double,
[tex]R_{2} = 2 R_{1}[/tex]
So we can write,
[tex]R_{1}^{2} \omega _{1} = 4R_{1}^{2} \omega _{2}[/tex]
For finding the angular velocity just before landing,
[tex]\omega _{2} = \frac{\omega _{1} }{4}[/tex]
[tex]\omega _{2} = \frac{20}{5}[/tex]
[tex]\omega _{2} = 5\frac{rad}{s}[/tex]
Therefore, the angular velocity just before landing on the trampoline bed is 5 [tex]\frac{rad}{s}[/tex]
