Answer:
[tex]\theta =1.14\ rev[/tex]
Explanation:
given,
θ = 1.5 revolution
initial angular speed = ωi = 3.25 rad/s
final angular speed = ωf = 1.60 rad/s
angular acceleration of the particle
[tex]\alpha = \dfrac{-\omega_i^2}{2\theta}[/tex]
[tex]\alpha = \dfrac{-3.25^2}{2\times 1.5 \times 2\pi\ rad}[/tex]
[tex]\alpha =-0.56\ rad/s^2[/tex]
Angular displacement of the wheel
using rotational formula to calculate displacement
[tex]\theta = \dfrac{\omega_f^2-\omega_i^2}{2\alpha}[/tex]
[tex]\theta = \dfrac{1.60^2-3.25^2}{2\times (-0.56)}[/tex]
[tex]\theta =7.14\ rad[/tex]
[tex]\theta =7.14\times \dfrac{1}{2\pi}[/tex]
[tex]\theta =1.14\ rev[/tex]
