Answer:
I = 16 kg*m²
Explanation:
Newton's second law for rotation
τ = I * α Formula (1)
where:
τ : It is the moment 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)
Kinematics of the wheel
Equation 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 (rad)
Data
ω₀ = 0
ωf = 1.2 rad/s
t = 2 s
Angular acceleration of the wheel
We replace data in the formula (2):
ωf = ω₀+ α*t
1.2= 0+ α*(2)
α*(2) = 1.2
α = 1.2 / 2
α = 0.6 rad/s²
Magnitude of the net torque (τ )
τ = F *R
Where:
F = tangential force (N)
R = radio (m)
τ = 80 N *0.12 m
τ = 9.6 N *m
Rotational inertia of the wheel
We replace data in the formula (1):
τ = I * α
9.6 = I *(0.6
)
I = 9.6 / (0.6
)
I = 16 kg*m²
Answer:
The center mass (Xcm) of the two mass = (M₁X₁ + M₂X₂)/(M₁ +M₂)
Explanation:
let the first mass = M
let the position of second = M
Potential energy is measured using formula Ep=mgh
m=mass (kg)
g= acceleration due to gravity (which is 9.8 on earth)
h= height in metres above ground
For this question
m=0.1
g=9.8
h=1
So Ep=0.1(9.8)(1)
Ep=0.98 Joules
When it is dropped all of this potential energy is converted into kinetic energy which can be measured using formula
Ek=1/2m(v^2) (v=final velocity)
Since all potential energy in this q is converted to kinetic we know Ek=0.98Joules and our mass is the same (0.1kg)
So when we sub everything in we get
0.98=1/2(0.1)(v^2)
0.98=0.05(v^2)||divide both side by 0.05
19.6=v^2 ||square root both sides
v=4.4 m/s
Center.........................
Explanation:
F = Gm1m2/r^2
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