At one end of a stick with length OA = R, we place a sphere. We push the sphere using a force of random magnitude so that it is rotated around point O (see Figure 6.6) creating an angle of Θ degrees, and the density function of this angle is given by ƒΘ(θ) = { 2/π, 0

Question

At one end of a stick with length OA = R, we place a sphere. We push the sphere using a force of random magnitude so that it is rotated around point O (see Figure 6.6) creating an angle of Θ degrees, and the density function of this angle is given by

[latex]f_\Theta ( heta )= \begin{cases} \frac{2}{\pi}, \quad 0 \lt heta \leq \frac{\pi}{2} , \ 0, \quad ~ ext{elsewhere}. \end{cases} [/latex]

Find the expected value of the position, (E(X), E(Y)), of the sphere after its rotation.

Accepted Answer

As can be seen from Figure 6.6, we have

X = R cos Θ, Y = R sin Θ,

so that we obtain, first for the expectation E(X),

[latex]E(X)=\int_{-\infty}^{\infty}R\cos heta f_{\Theta}( heta)\mathrm{d} heta=\int_{0}^{\pi/2}R\cos heta\cdot{\frac{2}{\pi}}\ \mathrm{d} heta={\frac{2R}{\pi}}\int_{0}^{\pi/2}\cos heta\mathrm{d} heta={\frac{2R}{\pi}}[\sin heta]_{0}^{\pi/2}={\frac{2R}{\pi}}\left(\sin{\frac{\pi}{2}}-\sin0 ight)={\frac{2R}{\pi}}.[/latex]

In a similar fashion, we find for E(Y) that

[latex]E(Y)=\int_{-\infty}^{\infty}R\sin heta f_{\Theta}( heta)\mathrm{d} heta=\int_{0}^{\pi/2}\!R\sin heta\cdot{\frac{2}{\pi}}\,\mathrm{d} heta={\frac{2R}{\pi}}\int_{0}^{\pi/2}\sin heta\,\mathrm{d} heta={\frac{2R}{\pi}}[-\cos heta]_{0}^{\pi/2}=-{\frac{2R}{\pi}}\left(\cos{\frac{\pi}{2}}-\cos0 ight)={\frac{2R}{\pi}}.[/latex]

Consequently, the expected position of the sphere lies on the angle bisector of the first quadrant of the Cartesian axes.

Question 6.8: At one end of a stick with length OA = R, we place a sphere.......

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