Answer:
a) -1.14 rev/min²
b) 9900 rev
c) -9.92×10⁻⁴ m/s²
d) 30.8 m/s²
Explanation:
First, convert hours to minutes:
2.2 h × 60 min/h = 132 min
a) Angular acceleration is change in angular velocity over change in time.
α = (ω − ω₀) / t
α = (0 rev/min − 150 rev/min) / 132 min
α = -1.14 rev/min²
b) θ = θ₀ + ω₀ t + ½ αt²
θ = 0 rev + (150 rev/min) (132 min) + ½ (-1.14 rev/min²) (132 min)²
θ = 9900 rev
c) The tangential component of linear acceleration is:
a_t = αr
First, convert α from rev/min² to rad/s²:
-1.14 rev/min² × (2π rad/rev) × (1 min / 60 s)² = -1.98×10⁻³ rad/s²
Therefore:
a_t = (-1.98×10⁻³ rad/s²) (0.50 m)
a_t = -9.92×10⁻⁴ m/s²
d) The magnitude of the net linear acceleration can be found from the tangential component and the radial component:
a² = (a_t)² + (a_r)²
The radial component is the centripetal acceleration:
a_r = v² / r
a_r = ω² r
First, convert 75 rev/min to rad/s:
75 rev/min × (2π rad/rev) × (1 min / 60 s) = 7.85 rad/s
Find the radial component:
a_r = (7.85 rad/s)² (0.50 m)
a_r = 30.8 m/s²
Now find the net linear acceleration:
a² = (-9.92×10⁻⁴ m/s²² + (30.8 m/s²)²
a = 30.8 m/s²
