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HW 9 Problem 10, Someone please help!!

0 votes
8,432 views

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.

image

(a) When switched on, the trimmer speeds up from 0 to 2 750 rev/min in 0.210 s. What average power is delivered to the head by the trimmer motor while it is accelerating?
(answer) W

(b) When the trimmer is cutting grass, it spins at 2 095 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?
(answer) W

asked Oct 18, 2013 in Intro College Physics by anonymous
recategorized Oct 19, 2013 by PhysicsDude

1 Answer

+1 vote

The key to solve this question is to calculate the moment of inertia of this trimmer. There are two components, one is the wire spool and the other is the plastic wire sticking out from the spool. We can use the following diagram to show how the trimmer can be considered in terms of components of simple structures/shapes. 

The next step is to find the masses of these different objects. The mass of the wires can be obtained by multiplying the length and the linear density (10g/m).

The mass of the trimmer spool is a bit challenging. First we treat this as uniform disks so we can get its area density of mass (defined as ||\lambda||) using the 0.1kg divided by the area of the circular ring: $$\lambda=\frac{0.1}{\pi R^2-\pi r^2}.$$

To find M, you multiple the area density with the area of the large disk ||M=\lambda \cdot \pi R^2||. 

Also you will have ||m=\lambda \cdot \pi R^2||.

Then you can calculate the total moment of inertia using ideas shown in the diagram. 

Once you have ||I_{total}||, you can find its final rotational kinetic energy ||K_{rot}=\frac {1}{2}I_{total}\omega ^2||.

The (b) part of the question is simpler. First you get the torque ||\tau=fL_1||. Since ||dW=\tau d\theta||, you can find: $$P=\frac{dW}{dt}=\tau \frac {d\theta}{dt}=\tau \omega .$$

 

answered Oct 19, 2013 by PhysicsDude (21,750 points)
edited Oct 19, 2013 by PhysicsDude
this is helpful, but what do we do after we solve for rotational kinetic energy in order to reach the power itself? Also how do we convert rev/min to angular velocity? do we account for the separate wire when we calculate total inertia? and is there a separate linear kinetic energy value for the extended wire?
Hi, is that wrong? m=M=λ⋅πR2?? Thx.
Angular Velocity in x(rev/min) = (x(rev/min)*2*pi) / 60 = x(rad/s)
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