The angular acceleration he must achieve to come to rest in 0.7s is -2.94 rad/s²
From the question,
The initial angular velocity of the gymnast = 2.06 rad/s
Also, he wishes to stop his angular motion in 0.7s,
That is, his final angular velocity would be 0 rad/s
To determine what angular acceleration he must achieve in this time to come to rest,
From one of the equations of kinematics for angular motion, we have that
[tex]\omega = \omega_{o} + \alpha t[/tex]
Where [tex]\omega[/tex] is the final angular velocity
[tex]\omega_{o}[/tex] is the initial angular velocity
[tex]\alpha[/tex] is the angular acceleration
and [tex]t[/tex] is time
From the question
[tex]\omega = 0 \ rad/s[/tex]
[tex]\omega_{o} = 2.06 \ rad/s[/tex]
[tex]t = 0.7\ s[/tex]
Putting these values into the equation, we get
[tex]0= 2.06 + \alpha \times 0.7[/tex]
[tex]0 = 2.06 + 0.7\alpha[/tex]
Then,
[tex]0.7\alpha = -2.06[/tex]
∴ [tex]\alpha = \frac{-2.06}{0.7}[/tex]
[tex]\alpha = -2.94 \ rad/s^{2}[/tex]
Hence, the angular acceleration he must achieve to come to rest in 0.7s is -2.94 rad/s²
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