Q. 25 - Calculus

Question

Use four slices to construct an approximation of each of the

quantities described in Exercises 25–30. In each case include

a labeled diagram and an explicit list of the values of $∆x$ and

each $x_k$ and ${x_k}^{*}$ (or $∆y$ and each $y_k$ and ${y_k}^{*}$ ).

The mass of a cylindrical rod with radius of 2 centimeters and length of 24 centimeters whose density at a point x centimeters from the left end is ρ(x) = 10.5 + 0.01527$x^2$ grams per cubic centimeter. 

Step-by-Step Solution

Answer

$$\begin{gathered} k_y=2.35ft \\ {k}_x=1.8ft \end{gathered}$$

Step1. Given Information

radius=2cm

length=24cm

density at left end = $\rho \left(x\right)=10.5+0.01527x^2\frac{gm}{c{m}^3}$

Step2. Calculate

$$\begin{gathered} dm=\rho dv=\rho \left(\pi x^2\right)dy \\ dm=\rho \pi \frac{1}{y^2}dy \\ d{I}_y=\frac{1}{2}dm{x}^2 \end{gathered}$$

Step3. Continue

$$\begin{gathered} d{I}_y=\frac{1}{2}\left(\rho \pi \frac{1}{y^2}dy\right)\left(\frac{1}{y^2}\right)=\frac{1}{2}\frac{\rho \pi dy}{y^4} \\ \int d{I}_y=\frac{1}{2}\rho \pi {\int }_{0.25}^4\frac{1}{y^4}dy = I_y=134+\rho \end{gathered}$$

Step4. Continue

$$\begin{gathered} c=\frac{1}{2}\left[\rho \pi \right(4^2\left)\right(0.25\left)\right]\left(4^2\right)=32\pi \rho \\ \int dm=m=\rho \pi {\int }_{0.25}^4\frac{1}{y^2}dy \Rightarrow m=24.34\rho \end{gathered}$$

Step5. Continue

$k_x=\sqrt{\frac{I_x}{m}}, k_y=\sqrt{\frac{I_y}{m}}$

Step6. Continue

$k_y=\sqrt{\frac{134\rho }{94.34\rho }} \Rightarrow k_y=2.35ft.$

Step7. Continue

$$\begin{gathered} d{I^{\text{'}}}_x=\frac{1}{4}dm{x}^2+dm{y}^2 \\ =\frac{1}{4}\left(\rho \pi \frac{dy}{y}\right)\frac{1}{y^2}+\left(\rho \pi \frac{dy}{y}\right)y^2=\rho \pi (\frac{1}{4y^4}+1)dy \\ \int d{I^{\text{'}}}_x={\int }_{0.25}^4\rho \pi (\frac{1}{4y^2}+1)dy \end{gathered}$$

Step8. Continue

${I^{\text{'}}}_x=(\frac{-1}{12y^3}+y){{|}_{0.25}}^4 \Rightarrow {I^{\text{'}}}_X=28.23y$

Step9. Continue

$$\begin{gathered} {I^{\text{'}\text{'}}}_x=\frac{1}{4}\left[\rho \pi \right(4^2\left)\right(0.25\left)\right]\left(4^2\right)+\left[\rho \pi \right(4^2\left)\right(0.25\left)\right](0.{125}^2) \\ {I^{"}}_x=50.46\rho \end{gathered}$$

Step10. Continue

$$\begin{gathered} I_x={I^{\text{'}}}_x+{I^{"}}_x =28.53\rho +50.46\rho \\ {I}_X=78.99\rho \\ {k}_x=\sqrt{\frac{78.99\rho }{24.37\rho }} \Rightarrow k_x=1.8ft. \end{gathered}$$