Respuesta :
Answer:
0.42 m/s²
Explanation:
r = radius of the flywheel = 0.300 m
w₀ = initial angular speed = 0 rad/s
w = final angular speed = ?
θ = angular displacement = 60 deg = 1.05 rad
α = angular acceleration = 0.6 rad/s²
Using the equation
w² = w₀² + 2 α θ
w² = 0² + 2 (0.6) (1.05)
w = 1.12 rad/s
Tangential acceleration is given as
[tex]a_{t}[/tex] = r α = (0.300) (0.6) = 0.18 m/s²
Radial acceleration is given as
[tex]a_{r}[/tex] = r w² = (0.300) (1.12)² = 0.38 m/s²
Magnitude of resultant acceleration is given as
[tex]a = \sqrt{a_{t}^{2} + a_{r}^{2}}[/tex]
[tex]a = \sqrt{0.18^{2} + 0.38^{2}}[/tex]
[tex]a[/tex] = 0.42 m/s²
