Answer:
(a) t = 22.9 s
(b) α= - 0.467 rad/s²
Explanation:
The uniformly accelerated circular movement, is a circular path movement in which the angular acceleration is constant.
We apply the equations of circular motion uniformly accelerated :
ωf ²= ω₀² + 2*α*θ Formula (1)
ωf= ω₀ + α*t Formula (2)
Where:
θ : angle that the body has rotated in a given time interval (rad)
α : angular acceleration (rad/s²)
t : time interval (s)
ω₀ : initial angular speed ( rad/s)
ωf : final angular speed ( rad/s)
Data
θ = 19.5 revolutions : angular displacement of each wheel or angle that the wheel has rotated in a given time interval
ω₀= 10.7 rad/s : initial angular speed of the Wheel ( rad/s)
ωf = 0 : final angular speed of the Whee( rad/s)
Calculating of the angular acceleration (α )
We replace data in the fómula (1),considering that 1 revolution is equal to 2π radians :
ωf ²= ω₀² + 2*α*θ
(0 )²= (10.7)² + 2*α*(19.5*2*π )
0= 114.49 + (245.04)*α
-114.49 = (245.04)*α
α= (-114.49) /(245.04)
α= -114.49 /(245.04)
α= -0.467 rad/s²
Time does it take for the bike to come to rest
We replace data in the formula (2)
ωf = ω₀ + α*t
0 = 10.7 + -0.467*t
-10.7 = - 0.467*t we multiply by (-1) both sides of the equation :
10.7 = 0.467*t
t = 10.7 / 0.467
t = 22.9 s
