Question

The head of a grass string trimmer has 100 gof cord wound in a light, cylindrical spool with inside diameter 3.00 cmand outside 18.0 cm diameteras shown in Figure P10.58. The cord has a linear density of$\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{g/m}}$. A single strand of the cord extends$\mathbf{16}\mathbf{}\mathbf{\text{cm}}$from the outer edge of the spool.

(a) When switched on, the trimmer speeds up from 0 to$\mathbf{0}\mathbf{-}\mathbf{2500}\mathbf{\text{\hspace{0.05em}rev/min}}$in0.215 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 $\mathbf{\text{2500\hspace{0.05em}rev/min}}$and the grass exerts an average tangential force of 7.65 Non the outer end of the cord, which is still at a radial distance of 16.0 cmfrom the outer edge of the spool. What is the power delivered to the head under load?

Short answer

$\begin{array}{rcl}\left(\mathrm{a}\right)P& =& 74.4\mathrm{W}\\ \left(\mathrm{b}\right)\mathrm{P}& =& 400\mathrm{W}\\ & & \end{array}$

Step-by-step solution

Given values

M ( mass of head of grass string trimmer) =$100 \mathrm{g}=0.1 \mathrm{kg}$.

$D_1$( inside diameter of the cylindrical spool )$=3 \text{cm}=0.03 \text{m}$.

$R_1$(inside radius of the cylindrical spool)$=1.5 \text{cm}=0.015\text{\hspace{0.05em}m}$.

$D_2$( outside diameter of the cylindrical spool )$=18 \text{cm}=0.18 \text{m}$.

$R_2$( outside radius of the cylindrical spool) $=9 \text{cm}=0.9 \text{m}$.

L (length of cord) $=16 \text{cm}=0.16 \text{m}$

$\begin{array}{c}\rho (\mathrm{liner}\mathrm{density}\mathrm{of}\mathrm{cord})=10 \mathrm{g}/\mathrm{m}\\ =0.01 \mathrm{kg}/\mathrm{m}\end{array}$

role="math" localid="1663733902383" $\begin{array}{c}{\omega }_1=2500\left(\frac{\mathrm{rev}}{\min }\right)\left(\frac{2\pi \mathrm{rad}}{1 \mathrm{rev}}\right)\left(\frac{1 \min }{60 \mathrm{s}}\right)\\ =262 \mathrm{rad}/\mathrm{s}\\ \begin{array}{c}{\omega }_2=2000\left(\frac{\mathrm{rev}}{\min }\right)\left(\frac{2\pi \mathrm{rad}}{1 \mathrm{rev}}\right)\left(\frac{1 \min }{60 \mathrm{s}}\right)\\ =209\mathrm{rad}/\mathrm{s}\end{array}\end{array}$

$\begin{array}{c}t=0.215 \mathrm{s}\\ F=7.65 \mathrm{N}\end{array}$

Find the average power

(a)

The moment of inertia of the cord wound in the cylindrical spool about its center of mass is given by Table {1 0 .2} as:

$I_{\mathrm{cylinder}}=\frac{1}{2}M\left(R_1^2+R_2^2\right)$

Substitute numerical values:

$\begin{array}{c}{I}_{\mathrm{cylinder}}=\frac{1}{2}(0.1)\left({0.015}^2+{0.09}^2\right)\\ =4.16\times {10}^{-4}\mathrm{kg}·{\mathrm{m}}^2\end{array}$

The moment of inertia of the extended strand (long, thin rod) with an axis of rotation through the center of mass is given by Table {1 0 . 2} as:

$I_{\mathrm{rod},\mathrm{CM}}=\frac{1}{12}m{L}^2$

Where (m) the mass of the single strand is given by:

$\begin{array}{c}m=\rho L\\ =(0.01)(0.16)\\ =1.6\times {10}^{-3} kg\end{array}$

Substitute for (m) from Equation (3) in Equation (2) and substitute numerical values:

$\begin{array}{c}{I}_{\mathrm{rod},\mathrm{CM}}=\frac{1}{12}\left(1.6\times {10}^{-3}\right)(0.16{)}^2\\ =3.41\times {10}^{-6}\mathrm{kg}·{\mathrm{m}}^2\end{array}$

According to the parallel-axis theorem, the moment of inertia of the rod about a parallel axis of rotation to the center of mass with an offset distance of ( 0.08 + 0.09 = 0.17 m) is:

$\begin{array}{c}{I}_{\mathrm{rod}}=I_{\mathrm{rod},\mathrm{CM}}+m(0.08+0.09{)}^2\\ =3.41\times {10}^{-6}+\left(1.6\times {10}^{-3}\right)(0.08+0.09{)}^2\\ =4.97\times {10}^{-5}\mathrm{kg}·{\mathrm{m}}^2\end{array}$

Hence, the total moment of inertia of the systemis:

$\begin{array}{rcl}I& =& I_{\mathrm{rod}}+I_{\mathrm{cylinder}}\\ I& =& 4.16\times {10}^{-4}+4.97\times {10}^{-5}\\ & =& 4.66\times {10}^{-4}\mathrm{kg}·{\mathrm{m}}^2\end{array}$

The average power delivered to the head by the trimmer motor is given by:

$\begin{array}{c}P=\frac{E}{\mathrm{\Delta }t}\\ P=\frac{\frac{1}{2}I{\omega }_1^2}{\mathrm{\Delta }t}\\ P=\frac{\frac{1}{2}\left(4.66\times {10}^{-4}\right){\left(262\right)}^2}{0.215}\\ =74.4\mathrm{W}\end{array}$

So, the average power delivered is 74.4 W.

(b)

The power delivered to the head under load is given by:

$\begin{array}{c}P=\tau {\omega }_2\\ =rF{\omega }_2\\ =(0.16+0.09)(7.65)\left(209\right)\\ =400\mathrm{W}\end{array}$

The power delivered to the head under load is 400 W.