Answer:
r = 20 m
Explanation:
The formula for the angular momentum of a rotating body is given as:
L = mvr
where,
L = Angular Momentum = 10000 kgm²/s
m = mass
v = speed = 2 m/s
r = radius of merry-go-round
Therefore,
10000 kg.m²/s = mr(2 m/s)
m r = (10000 kg.m²/s)/(2 m/s)
m r = 5000 kg.m ------------- equation 1
Now, the moment of inertia of a solid uniform disc about its axis through its center is given as:
I = (1/2) m r²
where,
I = moment of inertia = 50000 kg.m²
Therefore,
50000 kg.m² = (1/2)(m r)(r)
using equation 1, we get:
50000 kg.m² = (1/2)(5000 kg.m)(r)
(50000 kg.m²)/(2500 kg.m) = r
<u>r = 20 m</u>
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Answer:
1470kgm²
Explanation:
The formula for expressing the moment of inertial is expressed as;
I = 1/3mr²
m is the mass of the body
r is the radius
Since there are three rotor blades, the moment of inertia will be;
I = 3(1/3mr²)
I = mr²
Given
m = 120kg
r = 3.50m
Required
Moment of inertia
Substitute the given values and get I
I = 120(3.50)²
I = 120(12.25)
I = 1470kgm²
Hence the moment of inertial of the three rotor blades about the axis of rotation is 1470kgm²
Answer:
≈ 2.1 R
Explanation:
The moment of inertia of the bodies can be calculated by the equation
I = ∫ r² dm
For bodies with symmetry this tabulated, the moment of inertia of the center of mass
Sphere
= 2/5 M R²
Spherical shell
= 2/3 M R²
The parallel axes theorem allows us to calculate the moment of inertia with respect to different axes, without knowing the moment of inertia of the center of mass
I =
+ M D²
Where M is the mass of the body and D is the distance from the center of mass to the axis of rotation
Let's start with the spherical shell, axis is along a diameter
D = 2R
Ic =
+ M D²
Ic = 2/3 MR² + M (2R)²
Ic = M R² (2/3 + 4)
Ic = 14/3 M R²
The sphere
Is =
+ M [
²
Is = Ic
2/5 MR² + M
² = 14/3 MR²
² = R² (14/3 - 2/5)
= √ (R² (64/15)
= 2,066 R