Respuesta :
Answer:
The tension on the slack segment of the belt is 65 N
Explanation:
The given parameters are;
The mass of the solid disk = 70.5 kg
The diameter of the flywheel disk, D = 1.25 m
The radius of the pulley = 0.230 m
If the tension in the upper segment of the belt = 165 N
The angular acceleration of the flywheel = 1.67 rad/s²
The moment of inertia of a circular disc, I = 1/2·M·r²
The radius of the flywheel, r = D/2 = (1.25 m)/2 = 0.625 m
The moment of inertia of the flywheel, I = 1/2 × 70.5 kg × (0.625 m)² = 13.7695312 m²·kg
The moment of inertia, I = 13.7695312 m²·kg
Torque, τ = Moment of Inertia, I × Angular acceleration, α
Torque, τ = I × α
τ = 13.7695312 m²·kg × 1.67 rad/s² = 22.9951171 N·m ≈ 23 N·m
The torque applied from the pulley side is given as follows;
Let [tex]T_{slk}[/tex] represent the tension on the slack segment of the belts, we have;
23 N·m = (165 N - [tex]T_{slk}[/tex]) × 0.230 m
∴ 165 N - [tex]T_{slk}[/tex] = 23 N·m/0.230 m = 100 N
[tex]T_{slk}[/tex] = 165 N - 100 N = 65 N
The tension on the slack segment of the belt, [tex]T_{slk}[/tex] = 65 N
The tension on the slack segment of the belt is 65 N
What is flywheel?
flywheel, heavy wheel attached to a rotating shaft so as to smooth out delivery of power from a motor to a machine. The inertia of the flywheel opposes and moderates fluctuations in the speed of the engine and stores the excess energy for intermittent use.
The given parameters are;
The mass of the solid disk = 70.5 kg
The diameter of the flywheel disk, D = 1.25 m
The radius of the pulley = 0.230 m
If the tension in the upper segment of the belt = 165 N
The angular acceleration of the flywheel = 1.67 rad/s²
The moment of inertia of a circular disc, I = 1/2·M·r²
The radius of the flywheel, r = D/2 = (1.25 m)/2 = 0.625 m
Now we will calculate tension on the slack side
The moment of inertia of the flywheel,
[tex]I = \dfrac{1}{2} \times 70.5 kg \times (0.625 m)^2 = 13.7695312 \ m^2 \ kg[/tex]
The moment of inertia, I = 13.7695312 m²·kg
Torque, τ = Moment of Inertia, I × Angular acceleration, α
Torque, τ = I × α
τ = [tex]13.7695312 \times 1.67 = 22.9951171 Nm = 23 Nm[/tex]
The torque applied from the pulley side is given as follows;
Let represent the tension on the slack segment of the belts, we have;
23 N·m = (165 N - ) × 0.230 m
∴ 165 N - = 23 N·m/0.230 m = 100 N
= 165 N - 100 N = 65 N
The tension on the slack segment of the belt, = 65 N
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