The head of a grass string trimmer has 100 g of cord wound in a light, cylindric

Question

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 2 700 rev/min in 0.210 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 2 080 rev/min and the grass exerts an average tangential force of 6.35 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?
  W

Explanation / Answer

Given,

m = 100 g = 0.1 kg ; r1 = 3/2 = 1.5 cm = 0.015 m

R2 = 18/2 = 9 cm = 0.09 m ; m/l = 10 g/m = 0.01 kg/m ; x = 16 cm = 0.16 m

a)w = 2700 rev/min = 282.74 rad/s ; t = 0.21 s

Icord = 1/2 M (r1^2 + r2^2)

Ic = 0.5 x 0.1 x (0.015^2 + 0.09^2) = 4.16 x 10^-4 kg-m^2

density = mass/volume => mass = density x volume

mass = 0.01 x 0.16 = 1.6 x 10^-3 kg

I' = Icm + m d^2

I' = 1.6 x 10^-3 (1/2 x 0.16^2 + (0.09 + 0.08)^2) = 4.97 x 10^-5 kg-m^2

Itot = Ic + I'

Itot = 4.16 x 10^-4 + 4.97 x 10^-5 = 4.65 x 10^-4 kg m^2

KE = 1/2 I w^2

KE = 0.5 x 4.65 x 10^-4^2 x 282.74^2 = 18.59 J

From the defination of power

P = E/t = 18.59/0.21 = 88.52 W

Hence, P = 88.52 W

b)F = 6.35 N

w = 2080 rev/min = 217.82 rad/s

P = (Fd) w

P = 6.35 x (0.16 + 0.09)217.82 = 345.8 W

Hence, P = 345.8 W

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