A uniform helicopter rotor blade is $\mathbf{7}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{}\mathbf{\text{m}}\mathbf{}$long, has a mass of$110kg$ , and is attached to the rotor axle by a single bolt. (a) What is the magnitude of the force on the bolt from the axle when the rotor is turning at$320\text{rev}/\text{min}$? (Hint: For this calculation the blade can be considered to be a point mass at its center of mass. Why?) (b) Calculate the torque that must be applied to the rotor to bring it to full speed from rest in$6.70s$ . Ignore air resistance. (The blade cannot be considered to be a point mass for this calculation. Why not? Assume the mass distribution of a uniform thin rod.) (c) How much work does the torque do on the blade in order for the blade to reach a speed of $\mathbf{320}\mathbf{}\mathbf{}\mathbf{\text{rev}}\mathbf{/}\mathbf{\text{min}}$?

1. Length of rotor blade$\text{r}=7.80\text{m}$
2. Mass of rotor blade$\text{M}=110\text{kg}$
3. Angular frequency$\omega =320\text{rev}/\text{min}$
4. Time$\text{t}=6.70\text{s}$

Use the relation between angular velocity and the acceleration into the formula of force to find the magnitude of the force on the bolt from the axle when the rotor is rotating at$320\text{rev}/\text{min}$

. Then, use the parallel axis theorem to calculate the inertia. Using this inertia into the formula of torque, find the torque that must be applied to the rotor to bring it to full speed from rest in $6.70\text{s}$. Use the work energy principle in rotational motion to find the amount of work done by the torque on the blade in order for the blade to reach a speed of$320\text{rev}/\text{min}$

Formula:

$\text{F}=\text{ma}$

data-custom-editor="chemistry" $\text{a}={\omega }^2\text{r}$

$\tau =\text{I}\alpha$

$\text{I}={\text{I}}_{com}+{\text{md}}^2$

$\text{W}=\frac{1}{2}\text{I}{\omega }_f^2-\frac{1}{2}\text{I}{\omega }_i^2$

We have,

$\text{F}=\text{ma}$

but,

$\text{a}={\omega }^2\text{r}$

As

$\omega =\left(320\frac{\text{rev}}{\text{min}}\right)\left(\frac{1\text{min}}{60\text{sec}}\right)\left(\frac{2\pi \text{rad}}{1\text{rev}}\right)=33.49\frac{\text{rad}}{\text{s}}$

Therefore,

$\text{a}={\left(33.49\frac{\text{rad}}{\text{s}}\right)}^2\left(3.9\text{m}\right)$

$\Rightarrow \text{a}=4379.47\text{m}/{\text{s}}^2$

So,

$\text{F}=\left(110\text{kg}\right)\left(4379.47\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)$

$\Rightarrow \text{F}=4.81\times {10}^5\text{N}$

Therefore, the magnitude of the force on the bolt from the axel when the rotor is turning at$320\text{rev}/\text{min}$is $4.81\times {10}^5\text{N}$

For torque in terms of moment of inertia and angular acceleration,

$\tau =\text{I}\alpha$

But

$\text{I}={\text{I}}_{com}+{\text{md}}^2$

$$\begin{gathered} \Rightarrow \text{I}=\frac{{\text{ML}}^2}{12}+\text{M}{\left(\frac{\text{L}}{2}\right)}^2=\frac{{\text{ML}}^2}{12}+\frac{{\text{ML}}^2}{4} \\ \Rightarrow \text{I}=\frac{{\text{ML}}^2}{3} \\ \Rightarrow \text{I}=\frac{\left(110\text{kg}\right)(7.80\text{m}){}^2}{3} \\ \Rightarrow \text{I}=2230.8{\text{kg.m}}^2 \end{gathered}$$

And

$$\begin{gathered} \alpha =\frac{\omega }{\text{t}}=\frac{33.49\frac{\text{rad}}{\text{s}}}{9.7\text{s}} \\ \Rightarrow \alpha =4.999\text{rad}/{\text{s}}^2 \end{gathered}$$

Therefore,

$$\begin{gathered} \tau =\left(2230.8{\text{kg.m}}^2\right)\left(4.999\frac{\text{rad}}{{\text{s}}^{\text{2}}}\right) \\ \Rightarrow \tau =1.12\times {10}^4\text{NM} \end{gathered}$$

Therefore, the torque that must be applied to the rotor to bring it to full speed from rest in$6.70\text{s}$ is$1.12\times {10}^4\text{N.m}$

According to the work-energy principle for rotational motion work done is

$$\begin{gathered} \text{W}=\frac{1}{2}\text{I}{\omega }_f^2-\frac{1}{2}\text{I}{\omega }_i^2 \\ \Rightarrow \text{W}=\frac{1}{2}\left(2230.8{\text{kg.m}}^2\right){\left(33.49\frac{\text{rad}}{\text{s}}\right)}^2-\frac{1}{2}\left(2230.8{\text{kg.m}}^2\right)\left(0\frac{\text{rad}}{{\text{s}}^{\text{2}}}\right) \\ \Rightarrow \text{W}=1.25\times {10}^6\text{J}-0\text{J} \\ \Rightarrow \text{W}=1.25\times {10}^6\text{J} \end{gathered}$$

Therefore, the amount of work done by the torque on the blade in order for the blade to reach a speed of $320\text{rev}/\text{min}$is$1.25\times {10}^6\text{J}$