Question

Consider a square fan blade with sides of length 2 and the lower left corner placed at the origin. If the density of the blade is \(\rho (x,y) = 1 + 0.1x\), is it more difficult to rotate the blade about the \(x\)-axis or the \(y\)-axis?

Short answer

It is more difficult to rotate the fan about \(y\)-axis.

Step-by-step solution

Formula and concept used to find the direction in which it is difficult to rotate the blades of fan

The moments of inertia are obtained as following:

Here, the density function is given by \(\rho (x,y)\) and \(D\) is the region that is occupied by the lamina.

It is given that,

The region \(D\) is \(xy\)plane occupied by the fan blade with sides of length 2.

The density function is:

\(\rho (x,y) = 1 + 0.1x\).

From the given conditions it is observed that \(x\) varies from 0 to 2 and \(y\) varies from 0 to 2.

Obtain the direction of the fan blade in which it is difficult to rotate by find the moments of inertia about\(x\)and\(y\)axes.

Calculate the moment of inertia \({I_x}\)

Obtain the moment of inertia\({I_x}\) as follows:

Integrate with respect to \(y\) and apply the corresponding limit.

\(\begin{aligned}{c}{I_x} = \int_0^2 {\left( {\frac{{{y^3}}}{3} + \frac{{0.1x{y^3}}}{3}} \right)_0^2} dx\\ = \int_0^2 {\left( {\left( {\frac{{{{(2)}^3}}}{3} + \frac{{0.1x{{(2)}^3}}}{3}} \right) - \left( {\frac{{{{(0)}^3}}}{3} + \frac{{0.1x{{(0)}^3}}}{3}} \right)} \right)} dx\\ = \int_0^2 {\left( {\left( {\frac{8}{3} + \frac{{0.8x}}{3}} \right) - (0 + 0)} \right)} dx\\ = \int_0^2 {\left( {\frac{8}{3} + \frac{{0.8x}}{3}} \right)} dx\end{aligned}\)

Integrate with respect to \(x\) and apply the corresponding limit.

\(\begin{aligned}{c}{I_x} = \left( {\frac{{8x}}{3} + \frac{{0.8{x^2}}}{{3(2)}}} \right)_0^2\\ = \left( {\left( {\frac{{8(2)}}{3} + \frac{{0.8{{(2)}^2}}}{6}} \right) - \left( {\frac{{8(0)}}{3} + \frac{{0.8{{(0)}^2}}}{6}} \right)} \right)\\ = \left( {\left( {\frac{{16}}{3} + \frac{{0.8(4)}}{6}} \right) - (0 + 0)} \right)\\ = \left( {\frac{{16 + 1.6}}{3}} \right)\end{aligned}\)

On further simplification, the value of \({I_x}\) becomes:

\(\begin{aligned}{c}{I_x} = \frac{{17.6}}{3}\\ = 5.86666\\ \approx 5.87\end{aligned}\)

Calculate the moment of inertia \({I_y}\) 

Obtain the moment of inertia \({I_y}\) as follows:

Integrate with respect to \(y\) and apply the limit.

\(\begin{aligned}{c}{I_y} = \left. {\int_0^2 {\left( {{x^2} + 0.1{x^3}} \right)} (y)} \right)_0^2dx\\{I_y} = \int_0^2 {\left( {\left( {{x^2} + 0.1{x^3}} \right)(2 - 0)} \right)} dx\\ = \int_0^2 {\left( {\left( {{x^2} + 0.1{x^3}} \right)(2)} \right)} dx\\ = 2\int_0^2 {\left( {{x^2} + 0.1{x^3}} \right)} dx\end{aligned}\)

Integrate with respect to \(y\) and apply the limit.

\(\begin{aligned}{c}{I_y} = 2\left( {\frac{{{x^3}}}{3} + \frac{{0.1{x^4}}}{4}} \right)_0^2\\ = 2\left( {\left( {\frac{{{{(2)}^3}}}{3} + \frac{{0.1{{(2)}^4}}}{4}} \right) - \left( {\frac{{{{(0)}^3}}}{3} + \frac{{0.1{{(0)}^4}}}{4}} \right)} \right)\\ = 2\left( {\left( {\frac{8}{3} + \frac{{0.1(16)}}{4}} \right) - (0 + 0)} \right)\\ = 2\left( {\frac{8}{3} + \frac{{1.6}}{4}} \right)\end{aligned}\)

On further simplification, the value of \({I_y}\) becomes:

\(\begin{aligned}{c}{I_y} = 2\left( {\frac{{8(4) + 1.6(3)}}{{12}}} \right)\\ = 2\left( {\frac{{32 + 4.8}}{{12}}} \right)\\ = 2(3.07)\\ \approx 6.13\end{aligned}\)

Observe the moments of inertia and find the direction in which rotation of fan blade is more difficult

Here, \({I_x} < {I_y}\).

Therefore, rotation about \(y\)-axis is more difficult comparing to \(x\)-axis.

Thus, it is more difficult to rotate the fan about \(y\)-axis.