91Ó°ÊÓ

A slender rod of length \(80.0 \mathrm{~cm}\) and mass \(0.400 \mathrm{~kg}\) has its center of gravity at its geometrical center. But its density is not uniform; it increases by the same amount from the center of the rod out to either end. You want to determine the moment of inertia \(I_{\mathrm{cm}}\) of the rod for an axis perpendicular to the rod at its center, but you don't know its density as a function of distance along the rod, so you can't use an integration method to calculate \(I_{\mathrm{cm}}\). Therefore, you make the following measurements: You suspend the rod about an axis that is a distance \(d\) (measured in meters) above the center of the rod and measure the period \(T\) (measured in seconds) for small-amplitude oscillations about the axis. You repeat this for several values of \(d\). When you plot your data as \(T^{2}-4 \pi^{2} d / g\) versus \(1 / d\), the data lie close to a straight line that has slope \(0.320 \mathrm{~m} \cdot \mathrm{s}^{2} .\) What is the value of \(I_{\mathrm{cm}}\) for the rod?

Step by step solution

01

- Identify the oscillation formula

The period of oscillation of a physical pendulum, such as a suspended rod oscillating in small angles, is given by the formula \(T = 2\pi\sqrt{\frac{I_{\mathrm{cm}}}{mgd}}\). \(T\) is the period of oscillation, \(I_{cm}\) is the moment of inertia about the point of suspension, \(m\) is the mass of the pendulum, \(g\) is the acceleration due to gravity and \(d\) is the distance from the point of suspension to the center of gravity.

02

- Transform the oscillation formula

Rearrange the oscillation formula to isolate \(I_{cm}\) on one side, we get \(I_{cm} = \frac{T^2mgd}{4\pi^2}\). Given that when the term \(T^2 – 4\pi^2d/g\) (equivalent to \(\frac{4\pi^2I_{cm}}{mg}\)) is plotted against \(1/d\), the data form a straight line with slope 0.320 m.s².

03

- Calculate the moment of inertia

Equating the slope 0.320 m.s² with the derived term \(\frac{4\pi^2I_{cm}}{mg}\) from the oscillation formula, one can solve for \(I_{cm}\). Hence, \(I_{cm}\) = \(\frac{0.320 mg}{4\pi^2}\). Substituting the known values \(m = 0.400Kg\) and \(g = 9.8ms^{-2}\), the moment of inertia \(I_{cm} = 0.00818 Kg.m^2\).

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Key Concepts

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

Physical Pendulum

A physical pendulum is an extended object that oscillates around a pivot that is not at its center of mass, unlike the traditional simple pendulum which consists of a mass hanging on a string. This pivot creates a restoring torque as it swings back and forth around its equilibrium position.

For a slender rod, like the one in the exercise, its motion can be described as that of a physical pendulum. The inertia of the rod plays a significant role here. The moment of inertia, a measure of how the mass is distributed with respect to the pivot point, determines how the pendulum moves. The distribution of the rod’s mass affects how easily it swings, impacting the oscillation period.

• The formula for the oscillation period is derived using rotational motion equations.
• Large or asymmetric mass distributions increase inertia, slowing oscillations.

Understanding these fundamentals helps in analyzing how non-uniform density impacts motion in complex pendulums.

Oscillation Period

The oscillation period is the time it takes for the pendulum to complete one full back-and-forth swing. In our case, the formula for the oscillation period of a physical pendulum is given by:

\[ T = 2\pi\sqrt{\frac{I_{\mathrm{cm}}}{mgd}} \]

Here, \(T\) is the oscillation period, \(I_{\mathrm{cm}}\) is the moment of inertia at the center of mass (adjusted for the pivot point), \(m\) is mass, \(g\) is gravitational acceleration, and \(d\) is the distance from the pivot to the center of gravity.

• The equation tells us the period is influenced by inertia and mass distribution.
• Greater inertia leads to a longer period, indicating slower oscillations.

The challenge comes from a non-uniform density which makes calculating inertia typically complex. However, by measuring the period for known distances, one can infer inertia despite these complexities.

Non-uniform Density

Non-uniform density in a rod means that its mass is not evenly distributed from one end to the other. In our rod example, density increases towards each end, impacting both the center of gravity and the moment of inertia.

With non-uniform density, the center of mass doesn’t align with the geometric center, but in this exercise, symmetry ensures the center of gravity stays in the middle. This trait complicates the calculation of the moment of inertia using standard integration methods, as the mass distribution would need precise mathematical characterization.

• In non-uniform rods, density affects how resistance to rotation is distributed.
• Such unevenness results in different linear mass densities across the object.

Difficulty in analytically calculating inertia necessitates experimental methods as illustrated in the problem, using measured periods of oscillation.

Rod Mechanics

In physics, rod mechanics involves understanding how forces and motions interact along the length of a rod, especially when its properties, such as density or mass, are varied.

Our exercise focuses on this using a slender rod. The mechanical properties under scrutiny are the moment of inertia and how it affects oscillations. Because the rod has non-uniform density, its mechanical description isn't straightforward.

• Mechanical equilibrium and stability depend on how masses are distributed.
• Forces acting on a non-uniform rod lead to complex motion.

The physical pendulum formula applies here to derive measurable quantities, allowing moments of inertia calculation despite the intricacies introduced by the rod's heterogeneous density.