Question states: The head of a grass string trimmer has 100 g of cord wound in a
ID: 1285941 • Letter: Q
Question
Question states: 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 2 450 rev/min in 0.190 s. What average power is delivered to the head by the trimmer motor while it is accelerating?
(b) When the trimmer is cutting grass, it spins at 1 970 rev/min and the grass exerts an average tangential force of 8.00 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?
Explanation / Answer
(a)
The moment of inertia of the cord is,
I = (1/2)M[R12 + R22]
= (1/2)(0.100 kg)[(0.015 m)2 + (0.09 m)2]
= 0.00041625 kg.m2
The mass of the starnd is,
m = (10-2 kg/m)(0.16 m)
= 1.6x10-3 kg
Hence, the moment of inertia of the strand is,
I' = ICM + mr2
= 0.5mL2 + mr2
= (1.6x10-3 kg) [0.5(0.16 m)2 + [0.09 m + 0.08 m]2]
= 4.97x10-5 kg.m2
Hence, the total momnet of inertia is,
Itot = 0.00041625 kg.m2 + 4.97x10-5 kg.m2
= 4.66x10-4 kg.m2
The required power is,
P = E/t
= [0.5Itotw2]/t
= (0.5)(4.66x10-4 kg.m2)[(2450 rev/min)(2(pi) rad/60 s)]2 / (0.190 s)
= 80 W
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(b)
The required power is,
Torque = Fdw
= (8.00 N)(0.16 m + 0.09 m)(1970 rev/min)(2(pi) rad/60 s)
= 412.386 W
= 412.4 W (nearly)
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