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Chapter 10: Problem 51

A uniform rectangular flat plate has mass \(M\) and dimensions \(a\) by b. Use the parallel-axis theorem in conjunction with Table 10.2 to show that its rotational inertia about the side of length \(b\) is \(\frac{1}{3} M a^{2}\)

Short Answer

Expert verified
The rotational inertia of the uniform rectangular flat plate about the side of length \(b\) is \(\frac{1}{3} M a^{2}\)

Step by step solution

01

Understand and visualize the problem

The problem is presenting a rectangular flat plate with mass \(M\) and dimensions \(a\) and \(b\). The task is to find the rotational inertia or moment of inertia about the side of length \(b\).
02

Ready the parallel-axis theorem

The parallel-axis theorem is given by \(I = I_{cm} + M d^2\), where, \(I\) is the moment of inertia about some axis parallel to and a distance \(d\) away from the axis passing through the center of mass, \(I_{cm}\) is the moment of inertia of the body about the axis through the center of mass, and \(M\) is the mass of the body, and \(d\) is the distance between the axis.
03

Compute the moment of inertia about an axis through the center of mass

From Table 10.2, the center of mass of a thin rectangular flat plate is given by \(I_{cm} = \frac{1}{12} M (a^2 + b^2)\). As the axis is about the side of length \(b\), the distance \(d\) between this axis and the axis through the center of mass is \(a/2\).
04

Apply the parallel-axis theorem

Substituting for \(I_{cm}\), \(M\), and \(d\) in the expression of the parallel-axis theorem, we get \(I = \frac{1}{12} M (a^2 + b^2) + M (\frac{a}{2})^2\). Since the axis is through the side of length \(b\), \(b^2\) does not need to be considered in the computation of \(I_{cm}\). Thus, the equation simplifies as: \(I = \frac{1}{12} Ma^2 + \frac{1}{4} Ma^2 = \frac{1}{3} Ma^2\).

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Most popular questions from this chapter

Each propeller on a King Air twin-engine airplane consists of three blades, each of mass \(10 \mathrm{kg}\) and length \(125 \mathrm{cm} .\) The blades may be treated approximately as uniform, thin rods, (a) What's the propeller's rotational inertia? (b) If the plane's engine develops a torque of \(2.7 \mathrm{kN} \cdot \mathrm{m}\), how long will it take to spin up the propeller from 1400 rpm to 1900 rpm?

A \(320-\mathrm{N}\) frictional force acts on the rim of a 1.0 -m-diameter wheel to oppose its rotational motion. Find the torque about the wheel's central axis.

A solid sphere and a solid cube have the same mass, and the side of the cube is equal to the diameter of the sphere. The cube's rotation axis is perpendicular to two of its faces. Which has greater rotational inertia about an axis through the center of mass?

A ship's anchor weighs \(5000 \mathrm{N}\). Its cable passes over a roller of negligible mass and is wound around a hollow cylindrical drum of mass \(380 \mathrm{kg}\) and radius \(1.1 \mathrm{m},\) mounted on a frictionless axle. The anchor is released and drops \(16 \mathrm{m}\) to the water. Use energy considerations to determine the drum's rotation rate when the anchor hits the water. Neglect the cable's mass.

You have your bicycle upside-down for repairs. The front wheel is free to rotate and is perfectly balanced except for the \(25-g\) valve stem. If the valve stem is \(32 \mathrm{cm}\) from the rotation axis and at \(24^{\circ}\) below the horizontal, what's the resulting torque about the wheel's axis?
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