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A thin, uniform rod is bent into a square of side length \(a\). If the total mass is \(M\), find the moment of inertia about an axis through the center and perpendicular to the plane of the square. (\(Hint\): Use the parallel-axis theorem.)

Short Answer

Expert verified
The moment of inertia is \( \frac{5}{12}Ma^2 \).

Step by step solution

01

Understand the Problem

We have a rod bent into a square shape with sides of length \(a\). The total mass of the square is \(M\). We need to find the moment of inertia about an axis through the center and perpendicular to the plane of the square.
02

Determine Mass Distribution

Since the rod is thin and uniform, it has a uniform mass distribution. The total length of the rod when bent into a square is \(4a\). Thus, the mass per unit length is \(\frac{M}{4a}\). Hence, each side of the square, having length \(a\), has a mass \(m = \frac{Ma}{4a} = \frac{M}{4}\).
03

Use the Parallel-Axis Theorem

The parallel-axis theorem states: \(I = I_{cm} + md^2\), where \(I_{cm}\) is the moment of inertia about the center of mass and \(d\) is the distance from the axis passing through the center of mass to the rod's centerline.
04

Calculate Moment of Inertia for One Side

For a thin rod of mass \(m\) and length \(a\) about its center, the moment of inertia is \(I_{cm} = \frac{1}{12}ma^2\). For our square, each side can be considered a rod.
05

Apply to All Sides

For each side, the moment of inertia about the center axis perp to the plane is \(I_{side} = I_{cm} + md^2\), where \(d = \frac{a}{2}\) (half the side length to the center). For each side, this becomes \(I_{side} = \frac{1}{12}ma^2 + \frac{m(a/2)^2}{2} = \frac{1}{12}\frac{M}{4}a^2 + \frac{M}{4}(\frac{a}{2})^2 = \frac{5}{48}Ma^2\).
06

Sum All Sides

Since there are four sides making up the square, the total moment of inertia is \(I_{total} = 4 \times \frac{5}{48}Ma^2 = \frac{5}{12}Ma^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parallel-Axis Theorem
The parallel-axis theorem is a fundamental concept used in physics to calculate the moment of inertia for a body about any axis. This theorem is particularly useful when the axis of rotation does not pass through the center of mass. It enables us to calculate the moment of inertia of an object around an arbitrary axis if the moment of inertia about a parallel axis through the center of mass is known. The formula expressed by the theorem is:
  • \( I = I_{cm} + md^2 \)
Here, \( I \) is the moment of inertia about the new axis, \( I_{cm} \) is the moment of inertia about the center of mass, \( m \) is the mass of the object, and \( d \) is the perpendicular distance between the center of mass and the new axis. Each time we need to determine the moment of inertia about an axis not through the center of mass, this theorem provides a simple method to calculate that value.
Mass Distribution
Understanding the mass distribution of an object is critical when calculating the moment of inertia. In the given exercise, the rod that makes up the square is described as uniform, meaning that its mass is evenly distributed along its length. This uniform distribution implies each segment of the rod has the same mass per unit length.
This information is useful because it allows us to evaluate the mass of each section of the rod. The total mass of the rod is given as \( M \), and we know it forms a square, so its total length is \( 4a \). Therefore, the mass distribution per unit length is \( \frac{M}{4a} \). Consequently, each side of the square, which has a length \( a \), has a mass \( m = \frac{M}{4} \). Calculating this mass distribution is an essential step in determining how the mass of an object affects its inertia.
Uniform Rod
A uniform rod refers to a rod with a consistent distribution of mass along its length. This uniformity is one of the simplest and most fundamental examples in physics when studying mechanics, as it simplifies the calculations for various properties, including the moment of inertia.
For a rod bent into a different shape, such as a square in this exercise, the underlying concept of uniformity remains important. It denotes the same mass distribution throughout each segment. The moment of inertia for a uniform rod about its midpoint or center is given by the formula:
  • \( I_{cm} = \frac{1}{12}ma^2 \)
Where \( m \) is the mass of the rod and \( a \) is its length. As each side of the bent rod can be treated as an individual uniform rod, the simplicity aids us in extending our understanding to more complex shapes like the square.
Center of Mass
The center of mass is a vital concept when analyzing the dynamics of objects and systems. It is the point where the mass of the body or system is concentrated, and it acts as a balance point about which the mass is evenly distributed. For symmetric objects like a square formed from a uniform rod, the center of mass lies at the geometric center.
In the context of rotational motion, the center of mass is crucial because the simplest form of the moment of inertia calculation involves rotation about this point. In instances where the axis of rotation does not pass through the center of mass, the parallel-axis theorem becomes useful. Recognizing the center of mass helps in understanding how mass impacts motion and enables precise calculations of moments of inertia for complex-shaped objects.

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Most popular questions from this chapter

(a) What angle in radians is subtended by an arc 1.50 m long on the circumference of a circle of radius 2.50 m? What is this angle in degrees? (b) An arc 14.0 cm long on the circumference of a circle subtends an angle of 128 \(^\circ\). What is the radius of the circle? (c) The angle between two radii of a circle with radius 1.50 m is 0.700 rad. What length of arc is intercepted on the circumference of the circle by the two radii?

The angular velocity of a flywheel obeys the equation \(\omega_z\)(\(t\)) \(= A + Bt^2\), where \(t\) is in seconds and \(A\) and \(B\) are constants having numerical values 2.75 (for \(A\)) and 1.50 (for \(B\)). (a) What are the units of \(A\) and \(B\) if \(\omega_z\) is in rad/s? (b) What is the angular acceleration of the wheel at (i) \(t = 0\) and (ii) \(t =\) 5.00 s? (c) Through what angle does the flywheel turn during the first 2.00 s? (\(Hint\): See Section 2.6.)

How I Scales. If we multiply all the design dimensions of an object by a scaling factor \(f\), its volume and mass will be multiplied by \(f ^3\). (a) By what factor will its moment of inertia be multiplied? (b) If a \(\frac{1}{48}\)-scale model has a rotational kinetic energy of 2.5 J, what will be the kinetic energy for the full-scale object of the same material rotating at the same angular velocity?

A fan blade rotates with angular velocity given by \(\omega_z\)(\(t\)) \(= \gamma - \beta t^2\), where \(\gamma =\) 5.00 rad/s and \(\beta =\) 0.800 rad/s\(^3\). (a) Calculate the angular acceleration as a function of time. (b) Calculate the instantaneous angular acceleration \(\alpha_z\) at \(t =\) 3.00 s and the average angular acceleration \(\alpha_{av-z}\) for the time interval \(t =\) 0 to \(t =\) 3.00 s. How do these two quantities compare? If they are different, why?

A light, flexible rope is wrapped several times around a \(hollow\) cylinder, with a weight of 40.0 N and a radius of 0.25 m, that rotates without friction about a fixed horizontal axis. The cylinder is attached to the axle by spokes of a negligible moment of inertia. The cylinder is initially at rest. The free end of the rope is pulled with a constant force \(P\) for a distance of 5.00 m, at which point the end of the rope is moving at 6.00 m/s. If the rope does not slip on the cylinder, what is \(P\)?

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