The time required for this system to come to rest is equal to 9.87 seconds.
Given the following data:
In order to calculate the time required for this system to come to rest, we would have to determine the moment of inertia for gears A and C.
Mathematically, the moment of inertia for a gear can be calculated by using this formula:
I = mr²
Where:
For gear A, we have:
I = mr²
I = 0.675 × 0.04²
I = 0.675 × 0.0016
I = 1.08 × 10⁻³ kg·m².
For gear C, we have:
I = mr²
I = 3.6 × 0.1²
I = 3.6 × 0.01
I = 0.036 kg·m².
Next, we would convert the initial angular velocity of gear C in revolution per minutes (rpm) to radian per seconds (rad/s) as follows:
ωc₁ = 2000 × 2π/60
ωc₁ = 4000π/60
ωc₁ = 209.44 rad/s.
Also, the initial angular velocity of gears A and B is given by:
ωA₁ = ωB₁ = rc/rA × (ωc₁)
ωA₁ = ωB₁ = 0.15/0.06 × (209.44)
ωA₁ = ωB₁ = 2.5 × (209.44)
ωA₁ = ωB₁ = 523.60 rad/s.
Taking the moment about A, we have:
I_A·ωA₁ + rA∫F_{AC}dt - M(f)_A·t = 0
Substituting the given parameters into the formula, we have;
(1.08 × 10⁻³)·(523.60) + 0.06∫F_{AC}dt - 0.15t = 0
0.15t - 0.06∫F_{AC}dt = 0.56549 ......equation 1.
Similarly, the moment about B is given by:
0.15t - 0.06∫F_{BC}dt = 0.56549 ......equation 2.
Note: Let x = ∫F_{BC}dt + ∫F_{AC}dt
Adding eqn. 1 & eqn. 2, we have:
0.3t - 0.06x = (0.56549) × 2
0.3t - 0.06x = 1.13098 ......equation 3.
Taking the moment about A, we have:
Ic·ωc₁ - rC∫F_{AC}dt - rC∫F_{BC}dt - Mc(f)_A·t = 0
0.036(209.44) - 0.3t - 0.15(∫F_{BC}dt + ∫F_{AC}dt) = 0
0.3t + 0.15x = 7.5398 ......equation 4.
Solving eqn. 3 and eqn. 4 simultaneously, we have:
x = 30.5 Ns.
Time, t = 9.87 seconds.
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