Question

Calculate the moment of inertia of an align of point objects, as shown in Fig. 8–47 about (a) the y axis and (b) the x-axis. Assume two masses, \(m = 2.2\;{\rm{kg}}\)and \(M = 3.4\;{\rm{kg}}\), and the objects are wired together by very light, rigid pieces of wire. The align is rectangular and split through the middle by the x-axis. (c) About which axis would it be harder to accelerate this align?

FIGURE 8-47

Problem 39

Short answer

(a) The moment of inertia about the y-axis is \(7.0\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\).

(b) The moment of inertia about the x-axis is \(0.70\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\).

(c) It is harder to accelerate the mass system about the y-axis.

Step-by-step solution

Concept

The moment of inertia is the product of the mass and the square of the distance of the mass from the rotating axis. For this question, you have to find the x and y coordinates of each mass, and then you can find the moment of inertia about the x and y axes.

Given data

There are two \(m = 2.2\;{\rm{kg}}\) masses and two \(M = 3.4\;{\rm{kg}}\) masses.

The coordinate of the m mass in the upper left corner is \(\left( { - 0.50\;{\rm{m}},0.25\;{\rm{m}}} \right)\).

The coordinate of the m mass in the upper right corner is \(\left( {1.00\;{\rm{m}},0.25\;{\rm{m}}} \right)\).

The coordinate of the M mass in the lower-left corner is \(\left( { - 0.50\;{\rm{m}}, - 0.25\;{\rm{m}}} \right)\).

The coordinate of the M mass in the lower right corner is \(\left( {1.00\;{\rm{m}}, - 0.25\;{\rm{m}}} \right)\).

Calculation for part (a)

Part (a)

The moment of inertia about the y-axis is

\(\begin{align}{I_{\rm{y}}} &= \sum {{m_{\rm{i}}}x_{\rm{i}}^2} \\ &= m{\left( { - 0.50\;{\rm{m}}} \right)^2} + m{\left( {1.00\;{\rm{m}}} \right)^2} + M{\left( { - 0.50\;{\rm{m}}} \right)^2} + m{\left( {1.00\;{\rm{m}}} \right)^2}\\ &= \left( {2.2\;{\rm{kg}}} \right){\left( { - 0.50\;{\rm{m}}} \right)^2} + \left( {2.2\;{\rm{kg}}} \right){\left( {1.00\;{\rm{m}}} \right)^2} + 3.4\;{\rm{kg}}{\left( { - 0.50\;{\rm{m}}} \right)^2} + 3.4\;{\rm{kg}}{\left( {1.00\;{\rm{m}}} \right)^2}\\ &= 6.85\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}} \approx 7.0\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\end{align}\).

Hence, the moment of inertia about the y-axis is \(7.0\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\).

Calculation for part (b)

Part (b)

The moment of inertia about the x-axis is

\(\begin{align}{I_{\rm{x}}} &= \sum {{m_{\rm{i}}}y_{\rm{i}}^2} \\ &= m{\left( {0.25\;{\rm{m}}} \right)^2} + m{\left( {0.25\;{\rm{m}}} \right)^2} + M{\left( {0.25\;{\rm{m}}} \right)^2} + m{\left( {0.25\;{\rm{m}}} \right)^2}\\ &= \left( {2.2\;{\rm{kg}}} \right){\left( {0.25\;{\rm{m}}} \right)^2} + \left( {2.2\;{\rm{kg}}} \right){\left( {0.25\;{\rm{m}}} \right)^2} + 3.4\;{\rm{kg}}{\left( {0.25\;{\rm{m}}} \right)^2} + 3.4\;{\rm{kg}}{\left( {0.25\;{\rm{m}}} \right)^2}\\ &= 0.70\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\end{align}\).

Hence, the moment of inertia about the x-axis is \(0.70\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\).

Calculation for part (c)

Part (c)

When the value of the moment of inertia is larger, you need more torque to rotate the object.

The moment of inertia is bigger for the y-axis rotation.

Hence, it is harder to accelerate the mass system about the y-axis.