Physics, 26.10.2019 00:43, jennemylesp19oy5

The head of a grass string trimmer has 100 g of cord wound in a light, cylindrical spool with inside diameter 3.00 cm and outside diameter 18.0 cm as shown in the figure below. the cord has a linear density of 10.0 g/m. a single strand of the cord extends 16.0 cm from the outer edge of the spool. (a) when switched on, the trimmer speeds up from 0 to 2 250 rev/min in 0.230 s. what average power is delivered to the head by the trimmer motor while it is accelerating? w (b) when the trimmer is cutting grass, it spins at 1 915 rev/min and the grass exerts an average tangential force of 8.80 n on the outer end of the cord, which is still at a radial distance of 16.0 cm from the outer edge of the spool. what is the power delivered to the head under load?

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answered: iBrain

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;

$The \ average \ power =\dfrac{Rotational \ kinetic \ energy}{Time}$

The rotational kinetic energy = $E_{rotational } = \dfrac{1}{2} \cdot I \cdot \omega^2$

$\omega = \dfrac{ 2,250 \times 2\times \pi}{60} = 75$

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;

$I = \dfrac{1}{2} \cdot M \cdot (a^2 + b^2)$

Therefore, for the cord, we have;

$I = \dfrac{1}{2} \cdot M \cdot (r_1^2 + r_2^2)$

Therefore, we have;

$I_{cord} = \dfrac{1}{2} \times 0.1 \ kg\times ((0.09 \, m)^2 + (0.015 \, m)^2) = 0.00041625 \ kg/m^2$

$I_{cord}$ = 4.1625 × 10⁻⁴ kg/m²

The moment of inertia of the strand that extends from the cord is given as follows;

$I_{strand} = I_{cm} + M\cdot D^2$

Moment of inertia of cord

$I_{strand} = \dfrac{m \cdot L^2}{12} + m\cdot \left(\dfrac{L}{2} + r_2 \right)^2$

$I_{strand} = m \cdot \left(\dfrac{L^2}{12} + \left(\dfrac{L}{2} + r_2 \right)^2\right)$

$I_{strand} = \dfrac{10 \times 0.16}{1,000} \times \left(\dfrac{0.16^2}{12} + \left(\dfrac{0.16}{2} + 0.09 \right)^2\right) = \dfrac{115}{2316058} = 4.965\bar {3} \times 10^{-5}$

$I_{strand}$ = 4.965 $\bar 3$ kg/m²

The total moment of inertia, I = $I_{cord}$ + $I_{strand}$

∴ I = 4.1625 × 10⁻⁴ kg/m² + 4.965 $\bar 3$ kg/m² = 4.659 × 10⁻⁴ kg/m²

$E_{rotational } = \dfrac{1}{2} \times 4.659 \times 10^{-4}\cdot (75\times \pi)^2 \approx 12.933 \ J$

$Power = \dfrac{Energy}{Time}$

$Power\ delivered \ to \ head \approx \dfrac{12.933 \ J}{0.230 \ s} = 56.23 \ W$

(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

$\omega = \dfrac{ 1,915 \times 2\times \pi}{60} \approx 200.583$

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

Learn more about rolling torque here:

link

answered: Guest

magnityde and direction

answered: Guest

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

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The head of a grass string trimmer has 100 g of cord wound in a light, cylindrical spool with inside...