Solutions to Some Random Questions

Question 1:

The height of the center of mass is a mass-weighted sum of the coordinates of the individual centers of mass.

\[y_{CM} = \frac{y_1 M_1 + y_2 M_2 + y_3 M_3}{M_1 + M_2 + M_3}\]

Values of these quantities:

\[y_1 = 7R \hspace{1cm} M_1 = \frac{4}{3} \pi R^3 \rho_1 \\ y_2 = 4R \hspace{1cm} M_2 = \frac{32}{3} \pi R^3 \rho_2 \\ y_3 = R \hspace{1cm} M_3 = 8 R^3 \rho_3\]

Plugging in:

\[y_{CM} = \frac{7R \times \frac{4}{3} \pi R^3 \rho_1 + 4R \times \frac{32}{3} \pi R^3 \rho_2 + R \times 8 R^3 \rho_3}{\frac{4}{3} \pi R^3 \rho_1 + \frac{32}{3} \pi R^3 \rho_2 + 8 R^3 \rho_3 } = 4R \\ \frac{28}{3} \pi R^4 \rho_1 + \frac{128}{3} \pi R^4 \rho_2 + 8 R^4 \rho_3 = \frac{16}{3} \pi R^4 \rho_1 + \frac{128}{3} \pi R^4 \rho_2 + 32 R^4 \rho_3 \\ \frac{28}{3} \pi R^4 \rho_1 + 8 R^4 \rho_3 = \frac{16}{3} \pi R^4 \rho_1 + 32 R^4 \rho_3 \\ 4 \pi R^4 \rho_1 = 24 R^4 \rho_3 \\ \frac{\rho_1}{\rho_3} = \boxed{\frac{6}{\pi}} \]

Question 2:

The height of the center of mass is a length-weighted sum of the coordinates of the individual centers of mass of the straight and curved parts. We can use length instead of mass because of the uniform mass density per unit length.

\[y_{CM} = \frac{y_1 L_1 + y_2 L_2 }{L_1 + L_2 }\]

Straight Part:

\[L_1 = \pi R \hspace{1cm} y_1 = 0 \]

Curved Part:

\[L_2 = \pi R \hspace{1cm} y_2 = \frac{2R}{\pi} \]

Plugging in:

\[y_{CM} = \frac{0 \pi R + \frac{2R}{\pi} \pi R }{\pi R + \pi R} = \boxed{\frac{R}{\pi}}\]

Note by Steven Chase
6 months, 2 weeks ago

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Can you send the pdf of the paper and the name of the book in the background @Steven Chase

Ritik Agrawal - 6 months, 1 week ago

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I can't, because that image is all I have :)

Steven Chase - 6 months, 1 week ago

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Woah so cengage is famous even in USA?Btw which cengage book are you using?Is it the same as the one used by Indians(IIT advanced ones)?

Ayush Anand - 1 month, 4 weeks ago

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These questions were sent to me by somebody else.

Steven Chase - 1 month, 4 weeks ago

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oh alright.

Ayush Anand - 1 month, 3 weeks ago

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