Chapter 12: Q. 15 (page 330)

The three masses shown in FIGURE EX12.15 are connected by massless, rigid rods.
a. Find the coordinates of the center of mass.
b. Find the moment of inertia about an axis that passes through mass A and is perpendicular to the page.
c. Find the moment of inertia about an axis that passes through masses B and C.

Short Answer

Expert verified

The center of mass of the shown system. center of mass $(x, y) = (0.06, 0.04) m$

The moment of inertia about an axis that passes through mass A is $0.0020 kg m^{2}$

The moment of inertia about an axis that passes through masses B and C is $0.00128 kg m^{2}$

Step by step solution

Part (a)

The three masses connected by massless rigid rods is a rigid body.

The coordinate of center of mass is

$\begin{array}{l} x_{cm} = \frac{\sum m_{i} x_{i}}{\sum m_{i}} = \frac{m_{A} x_{A} + m_{B} x_{B} + m_{C} x_{C}}{m_{A} + m_{B} + m_{C}} \\ = \frac{(0.100 kg) (0 m) + (0.200 kg) (0.06 m) + (0.100 kg) (0.12 m)}{0.100 kg + 0.200 kg + 0.100 kg} = 0.060 m \\ y_{cm} = \frac{\sum m_{i} y_{i}}{\sum m_{i}} = \frac{m_{A} y_{A} + m_{B} y_{B} + m_{C} y_{C}}{m_{A} + m_{B} + m_{C}} \\ = \frac{(0.100 kg) (0 m) + (0.200 kg) (\sqrt{{(0.10 m)}^{2} - {(0.06 m)}^{2}}) + (0.100 kg) (0 m)}{0.100 kg + 0.200 kg + 0.100 kg} = 0.040 m \end{array}$

Part (b)

The moment of inertia about an axis through A and perpendicular to the page is

$\begin{array}{l} I_{A} = \sum m_{i} r_{i}^{2} = m_{B} {(0.10 m)}^{2} + m_{C} {(0.10 m)}^{2} \\ I_{A} = (0.100 kg) [{(0.10 m)}^{2} + {(0.10 m)}^{2}] \\ I_{A} = 0.0020 kg m^{2} \end{array}$

Part (c)

The distance of C from BC axis is $r_{A - B C} = \sqrt{{(0.10 m)}^{2} - {(0.06 m)}^{2}}$

The moment of inertia about an axis that passes through and is

$\begin{array}{l} I_{BC} = m_{A} {(r_{A - B C})}^{2} \\ I_{BC} = m_{A} {(\sqrt{{(0.10 m)}^{2} - {(0.06 m)}^{2}})}^{2} \\ I_{BC} = 0.00128 kg m^{2} \end{array}$

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Short Answer

Step by step solution

Part (a)

Part (b)

Part (c)

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