Answer:
[tex]\theta = 5.83\ rad[/tex]
Step-by-step explanation:
given,
angular deceleration, α = -0.5 rad/s²
final angular velocity,ω_f = 0 rad/s
angular position, θ = 6.1 rad
angular position at 3.9 s = ?
now, Calculating the initial angular speed
[tex]\omega_f^2 = \omega_i^2 + 2 \alpha \theta[/tex]
[tex] 0 = \omega_i^2 - 2\times 0.5\times 6.1[/tex]
[tex]\omega_i = \sqrt{6.1}[/tex]
[tex]\omega_i = 2.47\ rad/s[/tex]
now, angular position calculation at t=3.9 s
[tex]\theta = \omega_i t + \dfrac{1}{2}\alpha t^2[/tex]
[tex]\theta =2.47\times 3.9 - \dfrac{1}{2}\times 0.5\times 3.9^2[/tex]
[tex]\theta = 5.83\ rad[/tex]
Hence, the angular position of the wheel after 3.9 s is equal to 5.83 rad.
