Answer:
ωf = 2.19 rad/s
Explanation:
Newton's second law:
F = ma has the equivalent for rotation:
τ = I * α Formula (1)
where:
τ : It is the torque applied to the body. (Nxm)
I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)
α : It is angular acceleration. (rad/s²)
Data
F = 24 N : Tangential force
m : 14 kg : mass of the wheel
R = 0.38 m : radius of the wheel
Moment of inertia of the wheel
The moment of inertia of wheel is defined as follows:
I = m* R²
I = 14 kg*(0.38)²
I = 2.0216 kg*m²
Torque applied to the wheel
The Torque ( τ) applied to the wheel is defined as follows:
τ = F*d
Where:
F : Tangential force applied to the wheel
d : Perpendicular distance of the tangential force to the axis of rotation
τ = (24N)*(0.38 m)
τ = 9.12 N*m
Angular acceleration of the wheel (α
)
We replace data in the formula (1):
τ = I * α
9.12 = (2.0216) * α
α= 9.12 / (2.0216)
α = 4.5 rad/s²
Kinematics of the wheel
We apply the equations of circular motion uniformly accelerated :
ωf = ω₀ + α*t Formula (2)
Where:
α : Angular acceleration (rad/s²)
ω₀ : Initial angular speed ( rad/s)
ωf : Final angular speed ( rad
t : time interval (s)
Data
α = 4.5 rad/s²
ω₀ =1.6 rad/s
t = 0.13 s
We replace data in the formula (2):
ωf = ω₀ + α*t
ωf = 1.6 + (4.5)*(0.13)
ωf = 1.6 + (4.5)*(0.13)
ωf = 2.19 rad/s
