Answer:
A) α = -1.228 rev/min²
B) 7980 revolutions
C) α_t = -8.57 x 10^(-4) m/s²
D) α = 21.5 m/s²
Explanation:
A) Using first equation of motion, we have;
ω = ω_o + αt
Where,
ω_o is initial angular velocity
α is angular acceleration
t is time the flywheel take to slow down to rest.
We are given, ω_o = 140 rev/min ; t = 1.9 hours = 1.9 x 60 seconds = 114 s ; ω = 0 rev/min
Thus,
0 = 140 + 114α
α = -140/114
α = -1.228 rev/min²
B) the number of revolutions would be given by the equation of motion;
S = (ω_o)t + (1/2)αt²
S = 140(114) - (1/2)(1.228)(114)²
S ≈ 7980 revolutions
C) we want to find tangential component of the velocity with r = 40cm = 0.4m
We will need to convert the angular acceleration to rad/s²
Thus,
α = -1.228 x (2π/60²) = - 0.0021433 rad/s²
Now, formula for tangential acceleration is;
α_t = α x r
α_t = - 0.0021433 x 0.4
α_t = -8.57 x 10^(-4) m/s²
D) we are told that the angular velocity is now 70 rev/min.
Let's convert it to rad/s;
ω = 70 x (2π/60) = 7.33 rad/s
So, radial angular acceleration is;
α_r = ω²r = 7.33² x 0.4
α_r = 21.49 m/s²
Thus, magnitude of total linear acceleration is;
α = √((α_t)² + (α_r)²)
α = √((-8.57 x 10^(-4))² + (21.49)²)
α = √461.82
α = 21.5 m/s²
