Answer: [tex]933.76\ rad[/tex]
Explanation:
Given
The initial rate of gyroscope is [tex]\omega_o=34\ rad/s[/tex]
Angular deceleration [tex]\alpha =0.619\ rad/s^2[/tex]
Using angular equation of motion
[tex]\Rightarrow \omega^2-\omega_o^2=2\alpha \theta[/tex]
Insert the values
[tex]\Rightarrow 0-34^2=2(-0.619)\theta \\\\\Rightarrow \theta =\dfrac{34^2}{2\times 0.619}\\\\\Rightarrow \theta =933.76\ rad[/tex]
Thus, the angular displacement is [tex]933.76\ rad[/tex]
