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Chapter 11: Problem 39

Meter Stick Calculate the rotational inertia of a meter stick, with mass \(0.56 \mathrm{~kg}\), about an axis perpendicular to the stick and located at the \(20 \mathrm{~cm}\) mark. (Treat the stick as a thin rod.)

Short Answer

Expert verified
0.0971 kg m^2

Step by step solution

01

Identify the relevant formula

For a thin rod of length \(L\), mass \(m\) and an axis perpendicular to it at a distance \(d\) from the center of mass, the rotational inertia \(I\) is given by the formula: \[ I = I_{\text{cm}} + md^2 \] where \(I_{\text{cm}} = \frac{1}{12}mL^2\) is the inertia about the center of mass and \(d\) is the distance from the center of mass to the axis.
02

Plug in the given values

Given, \(m = 0.56 \mathrm{~kg}\), \(L = 1 \mathrm{~m}\) and the axis is at \(20 \mathrm{~cm}\) mark, which means \(d = 0.3 \mathrm{~m}\) (since center of mass of a meter stick is at the \(50 \mathrm{~cm}\) mark), we will now calculate \(I_{\text{cm}}\): \[ I_{\text{cm}} = \frac{1}{12} (0.56) (1^2) \mathrm{~kg} \mathrm{~m}^2 = \frac{0.56}{12} \mathrm{~kg} \mathrm{~m}^2 = 0.0467 \mathrm{~kg} \mathrm{~m}^2 \]
03

Calculate the distance term

Next, calculate the term \(md^2\): \[ md^2 = 0.56 \mathrm{~kg} \times (0.3 \mathrm{~m})^2 = 0.56 \times 0.09 \mathrm{~kg} \mathrm{~m}^2 = 0.0504 \mathrm{~kg} \mathrm{~m}^2 \]
04

Sum the contributions

Add \(I_{\text{cm}}\) and \(md^2\) to find the total rotational inertia: \[ I = I_{\text{cm}} + md^2 = 0.0467 \mathrm{~kg} \mathrm{~m}^2 + 0.0504 \mathrm{~kg} \mathrm{~m}^2 = 0.0971 \mathrm{~kg} \mathrm{~m}^2 \]

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Key Concepts

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

Rotational Motion
When an object spins around an axis, we describe this movement as rotational motion.
Unlike linear motion where objects move in a straight line, rotational motion involves the object rotating around a central point.
This is common in many real-world situations, such as a spinning bike wheel or a turning merry-go-round.
Key parameters to understand in rotational motion include angular velocity, angular acceleration, and rotational inertia.
These parameters help describe how quickly and smoothly an object spins.
Moment of Inertia
The moment of inertia, often referred to as rotational inertia, measures how much resistance an object offers to changes in its rotational motion.
It is analogous to mass in linear motion but applies to rotation. The moment of inertia depends on both the mass of the object and how this mass is distributed relative to the axis of rotation.
  • For simple shapes, such as rods and disks, specific formulas are used to calculate the moment of inertia.
  • The moment of inertia about the center of mass for a thin rod of length \(L\) and mass \(m\) is \( I_{\text{cm}} = \frac{1}{12}mL^2 \).
  • When the axis of rotation is not at the center of mass, we use the parallel axis theorem which states: \( I = I_{\text{cm}} + md^2 \) where \(d\) is the distance from the center of mass to the new axis.
These formulas simplify the process of determining rotational inertia for different objects and axis configurations.
Thin Rod
A thin rod is a common object often analyzed in physics problems involving rotational motion.
When the rod is rotated about an axis perpendicular to its length, we need to use specific formulas to find its moment of inertia.
For a rod of length \(1 \text{m}\) and mass \(0.56 \text{kg}\), rotating about an axis at \(20 \text{cm}\) from one end, we calculate the rotational inertia as follows:
  • First, determine the moment of inertia about the center of mass: \( I_{\text{cm}} = \frac{1}{12}mL^2 \) which gives \(0.0467 \text{kg} \text{m}^2\).
  • Next, find the distance from the center of mass to the rotation axis, here \(d = 0.3 \text{m}\).
  • Using the parallel axis theorem, calculate the term \(md^2 = 0.0504 \text{kg } \text{m}^2\).
  • Finally, sum these contributions to find the total rotational inertia: \( I = 0.0467 \text{kg} \text{m}^2 + 0.0504 \text{kg} \text{m}^2 = 0.0971 \text{kg} \text{m}^2 \).
This method can be applied to other scenarios to analyze the rotational properties of thin rods in different configurations.

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Most popular questions from this chapter

Bicycle Pedal Arm The length of a bicycle pedal arm is \(0.152 \mathrm{~m}\), and a downward force of \(111 \mathrm{~N}\) is applied to the pedal by the rider's foot. What is the magnitude of the torque about the pedal arm's pivot point when the arm makes an angle of (a) \(30^{\circ}\), (b) \(90^{\circ}\). and (c) \(180^{\circ}\) with the vertical?

A Disk A disk, initially rotating at \(120 \mathrm{rad} / \mathrm{s}\), is slowed down with a constant rotational acceleration of magnitude \(4.0 \mathrm{rad} / \mathrm{s}^{2} .(\mathrm{a})\) How much time does the disk take to stop? (b) Through what angle does the disk rotate during that time?

Polar Axis of Earth (a) What is the rotational speed \(\omega\) about the polar axis of a point on Earth's surface at a latitude of \(40^{\circ} \mathrm{N} ?\) (Earth rotates about that axis.) (b) What is the translational speed \(v\) of the point? What are (c) \(\omega\) and \((\mathrm{d}) v\) for a point at the equator?

Hands of a Clock What is the rotational speed of (a) the second hand, (b) the minute hand, and (c) the hour hand of a smoothly running analog watch? Answer in radians per second.

Four Identical Particles Four identical particles of mass \(0.50 \mathrm{~kg}\) each are placed at the vertices of a \(2.0 \mathrm{~m} \times 2.0 \mathrm{~m}\) square and held there by four massless rods, which form the sides of the square. What is the rotational inertia of this rigid body about an axis that (a) passes through the midpoints of opposite sides and lies in the plane of the square, (b) passes through the midpoint of one of the sides and is perpendicular to the plane of the square, and (c) lies in the plane of the square and passes through two diagonally opposite particles?
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