Respuesta :
Answer:
[tex]I=2.766\ kg.m^2[/tex]
Explanation:
We have:
diameter of the wheel, [tex]d=0.88\ m[/tex]
weight of the wheel, [tex]w_w=280\ N[/tex]
mass of hanging object to the wheel, [tex]m_o=6.32\ kg[/tex]
speed of the hanging mass after the descend, [tex]v_o=4\ m.s^{-1}[/tex]
height of descend, [tex]h=2.5\ m[/tex]
(a)
moment of inertia of wheel about its central axis:
[tex]I=\frac{1}{2} m.r^2[/tex]
[tex]I=\frac{1}{2} \frac{w_w}{g}.r^2[/tex]
[tex]I=\frac{1}{2} \times \frac{280}{9.8}\times 0.44^2[/tex]
[tex]I=2.766\ kg.m^2[/tex]
The moment of inertia of the wheel for an axis perpendicular to the wheel at its center is 17 kgm².
Acceleration of the mass
The acceleration of the mass is calculated as follows;
v² = u² + 2ad
v² = 0 + 2ad
v² = 2ad
a = v²/2d
a = (4²)/(2 x 2.5)
a = 3.2 m/s²
Angular acceleration of the wheel
α = a/R
where;
- R is the radius of the wheel = 0.44 m
α = 3.2/0.44
α = 7.27 rad/s²
Torque experienced by the wheel
The torque experienced by the wheel due to rope around it is calculated as follows;
τ = Fr = Iα
I = Fr/α
where;
- F is the applied force
- r is the radius of the wheel
- I is moment of inertia of the wheel
I = (mgr)/α
I = (280 x 0.44)/7.27
I = 17 kgm²
Thus, the moment of inertia of the wheel for an axis perpendicular to the wheel at its center is 17 kgm².
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