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Chapter 8: Q 39P (page 198)

Calculate the moment of inertia of an align of point objects, as shown in Fig. 8鈥�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

Expert verified

(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

01

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.

02

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 \(\left( {1.00\;{\rm{m}}, - 0.25\;{\rm{m}}} \right)\).

03

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 \(7.0\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\).

04

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 \(0.70\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\).

05

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.

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Most popular questions from this chapter

A merry-go-round accelerates from rest to\(0.68\;{\rm{rad/s}}\)in 34 s. Assuming the merry-go-round is a uniform disk of radius 7.0 m and mass 31,000 kg, calculate the net torque required to accelerate it.

The moment of inertia of a rotating solid disk about an axis through its CM is \(\frac{{\bf{1}}}{{\bf{2}}}{\bf{M}}{{\bf{R}}^{\bf{2}}}\) (Fig. 8鈥�20c). Suppose instead that a parallel axis of rotation passes through a point on the edge of the disk. Will the moment of inertia be the same, larger, or smaller? Explain why.

Question:(II) A figure skater can increase her spin rotation rate from an initial rate of 1.0 rev every 1.5 s to a final rate of 2.5 rev/s. If her initial moment of inertia was\(4.6\;{\rm{kg}} \cdot {{\rm{m}}^2}\), what is her final moment of inertia? How does she physically accomplish this change?

Two wheels having the same radius and mass rotate at the same angular velocity (Fig. 8鈥�38). One wheel is made with spokes so nearly all the mass is at the rim. The other is a solid disk. How do their rotational kinetic energies compare?

(a) They are nearly the same.

(b) The wheel with spokes has about twice the KE.

(c) The wheel with spokes has higher KE, but not twice as high.

(d) The solid wheel has about twice the KE.

(e) The solid wheel has higher KE, but not twice as high.

FIGURE 8-38

MisConceptual Question 7.

Two blocks are connected by a light string passing over a pulley of radius 0.15 m and the moment of inertia I. The blocks move (towards the right) with an acceleration of \({\bf{1}}{\bf{.00}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\) along their frictionless inclines (see Fig. 8鈥�51). (a) Draw free-body diagrams for each of the two blocks and the pulley. (b) Determine the tensions in the two parts of the string. (c) Find the net torque acting on the pulley, and determine its moment of inertia, I.

FIGURE 8-51

Problem 46
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