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Chapter 9: Problem 34

You are a project manager for a manufacturing company. One of the machine parts on the assembly line is a thin, uniform rod that is \(60.0 \mathrm{~cm}\) long and has mass \(0.400 \mathrm{~kg}\). (a) What is the moment of inertia of this rod for an axis at its center, perpendicular to the rod? (b) One of your engineers has proposed to reduce the moment of inertia by bending the rod at its center into a V-shape, with a \(60.0^{\circ}\) angle at its vertex. What would be the moment of inertia of this bent rod about an axis perpendicular to the plane of the \(\mathrm{V}\) at its vertex?

Short Answer

Expert verified
The moment of inertia for a straight rod is \(0.008 \mathrm{~kg} \cdot \mathrm{m}^2\) and for a V-shaped rod is \(0.006 \mathrm{~kg} \cdot \mathrm{m}^2\).

Step by step solution

01

Calculate moment of inertia for a straight rod

Use the formula for the moment of inertia of a straight rod which is \(I=\frac{1}{12}mL^2\) where \(m=0.4 \mathrm{~kg}\) is the mass and \(L=0.6 \mathrm{~m}\) is the length. You get \(I=\frac{1}{12}(0.4 \mathrm{~kg})(0.6 \mathrm{~m})^2=0.008 \mathrm{~kg} \cdot \mathrm{m}^2\).
02

Convert the angle in degrees to radians

In the second scenario, the angle is given in degrees, but we should convert it into radians as the trigonometric functions in the moment of inertia formula for a V-shaped rod use radians. The conversion formula from degrees to radians is \(\theta_{\text{rad}} = \frac{\pi}{180}\theta_{\text{deg}}\). Thus, we get \(\theta_{\text{rad}}=\frac{\pi}{180}(60.0^{\circ})=\frac{\pi}{3} \mathrm{~rad}\).
03

Calculate moment of inertia for a V-shaped rod

Now use the formula for the moment of inertia of a V-shaped rod which is \(I= \frac{mL^2}{4} sin^2(\frac{\theta}{2})\). We already have \( m=0.4 \mathrm{~kg}\), \(L=0.6 \mathrm{~m}\), and \(\theta=\frac{\pi}{3} \mathrm{~rad}\). Substituting these values into the formula, we get \( I= \frac{(0.4 \mathrm{~kg})(0.6 \mathrm{~m})^2}{4} sin^2(\frac{\pi}{6})=0.006 \mathrm{~kg} \cdot \mathrm{m}^2\).

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Key Concepts

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

Rigid Body Dynamics
When we talk about rigid body dynamics, we are exploring how solid objects move. Unlike fluids or gases, rigid bodies do not deform when forces are applied, at least in an ideal scenario. This makes them simpler to analyze. We often look at properties like mass, velocity, and acceleration, but a particularly important one in this context is the moment of inertia.
  • Rigid bodies tend to have fixed shapes and sizes.
  • Their mass does not change regardless of how they move.
  • We analyze such bodies by looking at how forces and torques affect them.
In our problem, the rod is considered a rigid body. The concept of the moment of inertia plays a crucial role in determining how it will react to rotational forces. It's like rotational mass - it measures how hard it is to make something spin around a particular axis. When we change the shape of the rod (like bending it into a V-shape), we change how that mass is distributed, thus changing the moment of inertia.
Simple Harmonic Motion
Simple harmonic motion (SHM) refers to a type of motion where an object moves back and forth through an equilibrium position. This is similar to the motion of a pendulum or a spring. While the moment of inertia isn鈥檛 directly about SHM, it's a fundamental element in broader mechanical physics which involves SHM.
  • SHM is characterized by periodic motion such as oscillations.
  • For something to have SHM, the restoring force must be proportional to the displacement.
  • The frequency and period of oscillation depend heavily on mass and how it is distributed.
In the context of our rod, if it were to oscillate back and forth in a motion similar to SHM, its moment of inertia would significantly influence the motion's frequency and amplitude. An altered shape, like bending the rod, might drastically change these oscillatory characteristics.
Mechanical Physics
Mechanical physics is a broad field that deals with the motion and behavior of objects. It encompasses concepts like force, energy, and motion, and is anchored on the laws of motion formulated by Isaac Newton.
  • It examines how bodies move in response to external forces.
  • Newton's laws serve as a foundation, describing how forces affect an object's movement.
  • Mechanical physics covers both linear and rotational movements.
In our exercise, the focus is on rotational dynamics, which falls under mechanical physics. Specifically, moment of inertia is a central theme. Understanding how a rod's shape or mass distribution affects its moment of inertia helps in controlling or predicting its rotational behavior. By bending the rod into a V-shape, the rules of mechanical physics allow us to predict the new moment of inertia and how it might behave under the same conditions of force or motion.

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Most popular questions from this chapter

An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev \(/\) min to 200 rev \(/ \min\) in 4.00 s. (a) Find the angular acceleration in rev/s \(^{2}\) and the number of revolutions made by the motor in the 4.00 s interval. (b) How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in part (a)?

The angular velocity of a flywheel obeys the equation \(\omega_{z}(t)=A+B t^{2},\) 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 \(\mathrm{rad} / \mathrm{s} ?\) (b) What is the angular acceleration of the wheel at (i) \(t=0\) and (ii) \(t=5.00 \mathrm{~s} ?\) (c) Through what angle does the flywheel turn during the first 2.00 s? (Hint: See Section \(2.6 .)\)

Measuring \(I\). As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis perpendicular to the wheel at its center. You measure the diameter of the wheel to be \(0.640 \mathrm{~m}\). Then you mount the wheel on frictionless bearings on a horizontal frictionless axle at the center of the wheel. You wrap a light rope around the wheel and hang an \(8.20 \mathrm{~kg}\) block of wood from the free end of the rope, as in Fig. E9.49. You release the system from rest and find that the block descends \(12.0 \mathrm{~m}\) in \(4.00 \mathrm{~s}\). What is the moment of inertia of the wheel for this axis?

\(9.88^{\circ}\) DATA You are analyzing the motion of a large flywheel that has radius \(0.800 \mathrm{~m}\). In one test run, the wheel starts from rest and turns in a horizontal plane with constant angular acceleration. An accelerometer on the rim of the flywheel measures the magnitude of the resultant acceleration \(a\) of a point on the rim of the flywheel as a function of the angle \(\theta-\theta_{0}\) through which the wheel has turned. You collect these results: $$ \begin{array}{l|cccccccc} \boldsymbol{\theta}-\boldsymbol{\theta}_{\mathbf{0}}(\mathbf{r a d}) & 0.50 & 1.00 & 1.50 & 2.00 & 2.50 & 3.00 & 3.50 & 4.00 \\ \hline \boldsymbol{a}\left(\mathbf{m} / \mathbf{s}^{\mathbf{2}}\right) & 0.678 & 1.07 & 1.52 & 1.98 & 2.45 & 2.92 & 3.39 & 3.87 \end{array} $$ Construct a graph of \(a^{2}\left(\right.\) in \(\left.\mathrm{m}^{2} / \mathrm{s}^{4}\right)\) versus \(\left(\theta-\theta_{0}\right)^{2}\) (in rad \(^{2}\) ). (a) What are the slope and \(y\) -intercept of the straight line that gives the best fit to the data? (b) Use the slope from part (a) to find the angular acceleration of the flywheel. (c) What is the linear speed of a point on the rim of the flywheel when the wheel has turned through an angle of \(135^{\circ} ?\) (d) When the flywheel has turned through an angle of \(90.0^{\circ},\) what is the angle between the linear velocity of a point on its rim and the resultant acceleration of that point?

A wheel of diameter \(40.0 \mathrm{~cm}\) starts from rest and rotates with a constant angular acceleration of \(3.00 \mathrm{rad} / \mathrm{s}^{2}\). Compute the radial acceleration of a point on the rim for the instant the wheel completes its second revolution from the relationship (a) \(a_{\mathrm{rad}}=\omega^{2} r\) and (b) \(a_{\mathrm{rad}}=v^{2} / r\)
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