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?

 

[HalfLostAndSearching]

Hello,

To find the average power delivered to the head by the trimmer motor while it is accelerating, we can use the equation P = Fv, where P is power, F is force, and v is velocity.

First, we need to find the force applied to the head of the trimmer. This can be found using the equation F = ma, where m is the mass and a is the acceleration. In this case, the mass is the linear density of the cord multiplied by the length of the cord that is extended, or 10.0 g/m * 0.16 m = 1.6 g.

Next, we need to find the acceleration. We know that the trimmer speeds up from 0 to 2 300 rev/min in 0.210 s, so we can convert this to radians per second by multiplying by 2π and dividing by 60, which gives us an angular velocity of 242.3 rad/s. The acceleration can then be found using the equation a = ωf^2, where ω is the angular velocity and f is the frequency. The frequency can be found by dividing the angular velocity by 2π, which gives us 38.5 Hz. Plugging these values into the equation, we get an acceleration of 2.33 x 10^5 m/s^2.

Now, we can plug the values for mass and acceleration into the equation F = ma to find the force, which is 3.73 x 10^-2 N.

Finally, we can use the equation P = Fv, where v is the velocity of the trimmer head. The velocity can be found using the equation v = ωr, where ω is the angular velocity and r is the radius of the spool. The radius can be found by subtracting the inner diameter from the outer diameter and dividing by 2, which gives us 7.5 cm or 0.075 m. Plugging in the values, we get a velocity of 18.2 m/s.

Now, we can plug the force and velocity values into the equation P = Fv to find the average power, which is 6.78 x 10^-1 watts or 0.678 watts.

I hope this helps! Let me know if you have any further questions.