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?

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 rotational kinetic energy =

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;

Therefore, for the cord, we have;

Therefore, we have;

= 4.1625 × 10⁻⁴ kg/m²

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

Moment of inertia of cord

= 4.965 kg/m²

The total moment of inertia, I = +

∴ I = 4.1625 × 10⁻⁴ kg/m² + 4.965 kg/m² = 4.659 × 10⁻⁴ kg/m²

(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

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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