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Chapter 9: Problem 67

A thin uniform rod 50.0 \(\mathrm{cm}\) long with mass 0.320 \(\mathrm{kg}\) is bent at its center into a \(\mathrm{V}\) shape, with a \(70.0^{\circ}\) angle at its vertex. Find the moment of inertia of this V-shaped object about an axis perpendicular to the plane of the \(\mathrm{V}\) at its vertex.

Short Answer

Expert verified
The moment of inertia is approximately \(0.0149 \, \text{kg} \, \text{m}^2\).

Step by step solution

01

Identify Moment of Inertia Formula

The moment of inertia \( I \) for a rod of length \( L \) and mass \( m \) about an axis through its center is given by the formula \( I = \frac{1}{12} m L^2 \). In this case, the rod is divided into two equal parts of \( 25 \text{ cm} \) each.
02

Calculate Mass of Each Rod Segment

Since the rod is uniform and bent at its center, each segment is \( 25 \text{ cm} \). Hence, the mass of each segment is \( \frac{0.320}{2} \ \mathrm{kg} = 0.160 \ \text{kg} \).
03

Moment of Inertia of Each Segment About Its Own Center

For each segment, the moment of inertia about its own center: \( I = \frac{1}{12} m L^2 = \frac{1}{12} \times 0.160 \times (0.25)^2 \). This calculates to approximately \( 0.000833 \ \text{kg} \, \text{m}^2 \) for each segment.
04

Apply the Parallel Axis Theorem

The Parallel Axis Theorem states \( I = I_{cm} + md^2 \) where \( d \) is the perpendicular distance from the centroid to the axis. Here, \( d = \frac{L}{2}\cos(\theta / 2) \). So \( d = 0.25 \times \cos(35^\circ) \approx 0.204 \ \text{m} \).
05

Calculate Each Segment's Inertia About Vertex Axis

Inertia of each segment about the vertex using the parallel axis theorem: \( I_{segment} = I + md^2 = 0.000833 + 0.160 \times 0.204^2 \). This results in approximately \( 0.007467 \ \text{kg} \, \text{m}^2 \) per segment.
06

Total Moment of Inertia for the V-shape

Since there are two such segments, the total moment of inertia \( I_{total} = 2 \times 0.007467 \ \text{kg} \, \text{m}^2 = 0.014934 \ \text{kg} \, \text{m}^2 \).

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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 pivotal concept in rotational dynamics. It helps us find the moment of inertia of an object around any axis, provided we know the moment of inertia around a parallel axis that passes through the object's center of mass (COM). Think of it as a handy tool for shifting the axis of rotation without recalculating from scratch.

The theorem mathematically states that the moment of inertia about any axis is the sum of its moment of inertia about a parallel axis through the center of mass and the product of its mass and the square of the perpendicular distance between the two axes. In formula terms, this is expressed as:
  • \( I = I_{cm} + md^2 \),
where:
  • \( I \) is the moment of inertia about the new axis,
  • \( I_{cm} \) is the moment of inertia about the center of mass axis,
  • \( m \) is the total mass of the object,
  • \( d \) is the distance between the two parallel axes.
Utilizing this theorem, you can calculate the inertia for more complex shapes like the V-shaped rod by treating each segment separately and then combining their results.
Centre of Mass
The centre of mass (COM) is the specific point where the entire mass of an object is imagined to be concentrated for calculations pertaining to translational and rotational motion.

For a uniformly distributed object such as a rod, its center of mass is typically in the middle. When a rod is bent, things become a tad more complex. Each section might have its own center of mass when bent, making it essential to analyze each segment separately, as they affect how the object rotates.
  • In this V-shaped rod scenario, the COM for rotational axis considerations intrigues us more than the COM of the entire object, as we consider rotational dynamics about the vertex.
Positioning your axis at the COM simplifies calculations and is crucial in applying concepts like the Parallel Axis Theorem effectively.
Rotational Dynamics
Rotational dynamics, distinct from linear dynamics, examines how objects spin or rotate. Just like linear dynamics relies heavily on concepts of mass and acceleration, rotational dynamics pivots on moment of inertia and angular acceleration.

To understand rotational dynamics, appreciate how moment of inertia plays the role that mass does in linear motion鈥攊t is a measure of an object's resistance to change in its rotational motion.
  • A higher moment of inertia means more effort is required to change the object's spinning speed.
  • It's not just about mass鈥攈ow mass is distributed relative to the pivot point matters too.
      • In rotating systems, angular velocity, angular acceleration, and torque work collaboratively, akin to velocity, acceleration, and force in linear systems.
      • Equations like \( \tau = I \alpha \) (torque equals moment of inertia times angular acceleration) play a key role.

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Most popular questions from this chapter

A A flywheel having constant angular acceleration requires 4.00 s to rotate through 162 rad. Its angular velocity at the end of this time is 108 rad/s. Find (a) the angular velocity at the beginning of the 4.00 s interval; (b) the angular acceleration of the flywheel.

Storing energy in flywheels. It has been suggested that we should use our power plants to generate energy in the off-hours (such as late at night) and store it for use during the day. One idea put forward is to store the energy in large fly wheels. Suppose we want to build such a flywheel in the shape of a hollow cylinder of inner radius 0.500 \(\mathrm{m}\) and outer radius \(1.50 \mathrm{m},\) using concrete of density \(2.20 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} .\) (a) If, for stability, such a heavy flywheel is limited to 1.75 second for each revolution and has negligible friction at its axle, what must be its length to store 2.5 \(\mathrm{MJ}\) of energy in its rotational motion? (b) Suppose that by strengthening the frame you could safely double the flywheel's rate of spin. What length of flywheel would you need in that case? (Solve this part without reworking the entire problem!)

\(\cdot\) A car is traveling at a speed of 63 \(\mathrm{mi} / \mathrm{h}\) on a freeway. If its tires have diameter 24.0 in and are rolling without sliding or slipping, what is their angular velocity?

\(\bullet\) A vacuum cleaner belt is looped over a shaft of radius 0.45 \(\mathrm{cm}\) and a wheel of radius 2.00 \(\mathrm{cm} .\) The motor turns the shaft at 60.0 \(\mathrm{rev} / \mathrm{s}\) and the moving belt turns the wheel, which in turn is connected by another shaft to the roller that beats the dirt out of the rug being vacuumed (see Figure 9.32 ). Assume that the belt doesn't slip on either the shaft or the wheel. (a) What is the speed of a point on the belt? (b) What is the angular velocity of the wheel, in rad/s?

\(\cdot\) A turntable that spins at a constant 78.0 rpm takes 3.50 s to reach this angular speed after it is turned on. Find (a) its angular acceleration (in \(\mathrm{rad} / \mathrm{s}^{2} ),\) assuming it to be constant, and (b) the number of degrees it turns through while speeding up.
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