Answer: [tex]\alpha=4.798\ rad/s^2\quad [\text{deceleration}][/tex]
Explanation:
Given
Initial revolution of flywheel [tex]N_1=600\ rpm[/tex]
(Initial angular velocity [tex]\omega_i=\dfrac{2\pi N_1}{60}[/tex])
Final revolution of flywheel [tex]N_2=600\ rpm[/tex]
(Final angular velocity [tex]\omega_f=\dfrac{2\pi N_1}{60}[/tex])
Revolution turned [tex]34[/tex]
So, angle turned is [tex]\theta =2\pi \times 34\\[/tex]
Using equation of angular motion i.e. [tex]\omega_f^2-\omega_i^2=2\cdot \alpha \cdot \theta[/tex]
[tex]\Rightarrow \left(\dfrac{2\pi \times 416}{60}\right)^2-\left(\dfrac{2\pi \times 416}{60}\right)^2=2\alpha \times (68\pi )\\\\\Rightarrow \alpha=\dfrac{1898.344-3948.86}{427.312}\\\\\Rightarrow \alpha =-4.798\ rad/s^2\\\Rightarrow \alpha =4.798\ rad/s^2\quad [\text{deceleration}][/tex]
