How Much Power Does a Trimmer Motor Deliver During Acceleration?

[MD_Programmer]

Homework Statement

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.

http://www.webassign.net/pse/p10-50.gif

When switched on, the trimmer speeds up from 0 to 2 300 rev/min in 0.210 s. What average power is delivered to the head by the trimmer motor while it is accelerating?

Homework Equations

I = Icm + MD^2
1/2 m (R1^2 + R2^2)

P = 1/2 I ωf^2

The Attempt at a Solution

can't figure out a exact one :(

please help explain the work you did to find the answer because I can't get this one at all

 

[Simon Bridge]

The average power is the total work done divided by the time it took.

 

[MD_Programmer]

that is true but this is the formula i derived but i hope its right

I = m*[L^2/12 + (R2 + L/2)^2]

then plug that into power formula

P = 1/2* I * ωf^2/ Δt

the issue is what should L in both parts of the I equation? I know R2 is 0.17 m / 2 to get .085 radius

 

[Simon Bridge]

The equations are meaningless without the reasoning.

What is R2 supposed to represent. What is L supposed to represent?
Some radius and some angular momentum sure - but of what? Or is L some length?

It helps to troubleshoot your equations if you are clear about what each of the letters means.
It's a good discipline for you when you troubleshoot yourself too and it is worth marks in an exam.

Lets see... You will certainly need the moment of inertia of the trimmer-head...
How are you modelling that? You've not said.

Guessing: Looks like you are modelling it as just a rod length L - rotating about a pivot displaced from com by the outer radius of the hoop and half the length of the rod. Is that all there is to the trimmer head?