Answer :
The rotational equivalence of a force is a torque
(a) Power delivered to the head is approximately 56.23 W
(b) Power delivered to the head under load is approximately 441.184 W
The reason why the above values are correct is as follows;
The known parameters are;
Mass of the cord, m = 100 g
Inside diameter of the cord, D₁ = 3.00 cm (∴ r₁ = 0.015 m)
Outside diameter of the cord, D₂ = 18.00 cm (∴ r₂ = 0.09 m)
Linear density of the cord = 10.0 g/m
The length of a strand of cord, L = 16.0 cm
(a) Time it takes the trimmer to speed up to 2,250 rev/min from 0 is 0.230 s.
Required;
The average power delivered to the head by the trimmer motor while it is accelerating
Solution;
[tex]The \ average \ power =\dfrac{Rotational \ kinetic \ energy}{Time}[/tex]
The rotational kinetic energy = [tex]E_{rotational } = \dfrac{1}{2} \cdot I \cdot \omega^2[/tex]
[tex]\omega = \dfrac{ 2,250 \times 2\times \pi}{60} = 75[/tex]
Angular speed, ω = 75 rad/s
I = Moment of inertia of the cord + Moment of inertia of strand
The cord is a hollow cylinder
Moment of inertia, I, of a hollow cylinder, with radii, a and b, is given as follows;
[tex]I = \dfrac{1}{2} \cdot M \cdot (a^2 + b^2)[/tex]
Therefore, for the cord, we have;
[tex]I = \dfrac{1}{2} \cdot M \cdot (r_1^2 + r_2^2)[/tex]
Therefore, we have;
[tex]I_{cord} = \dfrac{1}{2} \times 0.1 \ kg\times ((0.09 \, m)^2 + (0.015 \, m)^2) = 0.00041625 \ kg/m^2[/tex]
[tex]I_{cord}[/tex] = 4.1625 × 10⁻⁴ kg/m²
The moment of inertia of the strand that extends from the cord is given as follows;
[tex]I_{strand} = I_{cm} + M\cdot D^2[/tex]
Moment of inertia of cord
[tex]I_{strand} = \dfrac{m \cdot L^2}{12} + m\cdot \left(\dfrac{L}{2} + r_2 \right)^2[/tex]
[tex]I_{strand} = m \cdot \left(\dfrac{L^2}{12} + \left(\dfrac{L}{2} + r_2 \right)^2\right)[/tex]
[tex]I_{strand} = \dfrac{10 \times 0.16}{1,000} \times \left(\dfrac{0.16^2}{12} + \left(\dfrac{0.16}{2} + 0.09 \right)^2\right) = \dfrac{115}{2316058} = 4.965\bar {3} \times 10^{-5}[/tex]
[tex]I_{strand}[/tex] = 4.965[tex]\bar 3[/tex] kg/m²
The total moment of inertia, I = [tex]I_{cord}[/tex] + [tex]I_{strand}[/tex]
∴ I = 4.1625 × 10⁻⁴ kg/m² + 4.965[tex]\bar 3[/tex] kg/m² = 4.659 × 10⁻⁴ kg/m²
[tex]E_{rotational } = \dfrac{1}{2} \times 4.659 \times 10^{-4}\cdot (75\times \pi)^2 \approx 12.933 \ J[/tex]
[tex]Power = \dfrac{Energy}{Time}[/tex]
[tex]Power\ delivered \ to \ head \approx \dfrac{12.933 \ J}{0.230 \ s} = 56.23 \ W[/tex]
(b) The rotational speed under load = 1,915 rev/min
The tangential force exerted on the outer end of the cord, F = 8.80 N
The distance of the cord from the spool = 16.0 cm
Torque = F × d
d = 0.16 m + 0.09 m = 0.25 m
Torque = 8.80 N × 0.25 m = 2.2 N·m
Power = Torque × ω
[tex]\omega = \dfrac{ 1,915 \times 2\times \pi}{60} \approx 200.583[/tex]
ω ≈ 200.583 rad/s
Power = 2.2 N·m × 200.583 rad ≈ 441.184 W
Power delivered to the head under load ≈ 441.184 W
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