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Chapter 11: Problem 92

A thin meter stick of mass \(150 \mathrm{g}\) rotates around an axis perpendicular to the stick's long axis at an angular velocity of 240 rev/min. What is the angular momentum of the stick if the rotation axis (a) passes through the center of the stick? (b) Passes through one end of the stick?

Short Answer

Expert verified
The angular momentum of the stick when (a) the rotation axis passes through the center is \(0.2\pi\,\text{kg m}^{2}\text{/s}\), and when (b) the rotation axis passes through one end is \(0.8\pi\,\text{kg m}^{2}\text{/s}\).

Step by step solution

01

Convert angular velocity to rad/s

The angular velocity is given as 240 rev/min. We need to convert it to rad/s. To do this, we'll use the conversion factors: \(1 \,\text{rev} = 2\pi\,\text{rad}\) and \(1 \,\text{min} = 60 \,\text{s}\). \( \omega = 240 \,\text{rev/min} \times 2\pi\,\text{rad/rev} \times \frac{1\,\text{min}}{60\,\text{s}} = 16\pi\,\text{rad/s}\)
02

Find moment of inertia for case (a)

For case (a) where the axis passes through the center of the stick, a thin rod has the following moment of inertia: \(I = \frac{1}{12}\times M \times L^{2}\), where M is the mass of the rod and L is its length. Given, mass \(M = 150 \,\text{g} = 0.15\,\text{kg}\) and length \(L = 1\,\text{m}\). Therefore, \(I_{a} = \frac{1}{12}\times 0.15\,\text{kg}\times (1\,\text{m})^{2} = 0.0125\,\text{kg m}^{2}\).
03

Find angular momentum for case (a)

For case (a), we can find the angular momentum using the formula: \(L = I\omega\), where I is the moment of inertia and 蠅 is the angular velocity. So, \(L_{a} = I_{a}\times \omega = 0.0125\,\text{kg m}^{2}\times 16\pi\,\text{rad/s} = 0.2\pi\,\text{kg m}^{2}\text{/s}\).
04

Find moment of inertia for case (b)

For case (b) where the axis passes through one end of the stick, a thin rod has the following moment of inertia: \(I = \frac{1}{3}\times M \times L^{2}\). Using the same values for M and L as before, we get: \(I_{b} = \frac{1}{3}\times 0.15\,\text{kg}\times (1\,\text{m})^{2} = 0.05\,\text{kg m}^{2}\).
05

Find angular momentum for case (b)

For case (b), again, we can find the angular momentum using the formula: \(L = I\omega\). So, \(L_{b} = I_{b}\times \omega = 0.05\,\text{kg m}^{2}\times 16\pi\,\text{rad/s} = 0.8\pi\,\text{kg m}^{2}\text{/s}\). Now we have the angular momentum for both cases: (a) When the rotation axis passes through the center of the stick, the angular momentum is \(0.2\pi\,\text{kg m}^{2}\text{/s}\). (b) When the rotation axis passes through one end of the stick, the angular momentum is \(0.8\pi\,\text{kg m}^{2}\text{/s}\).

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Key Concepts

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

Moment of Inertia
The moment of inertia is a critical concept when dealing with rotating objects. It quantifies how difficult it is to change the rotational motion of an object. This concept can be thought of as the rotational equivalent of mass in linear motion. To better understand, consider a simple example: a spinning figure skater. When the skater pulls their arms in, they spin faster. This change in spin is due to the change in the moment of inertia. When the arms are extended, the moment of inertia increases, and when the arms are pulled in, it decreases. For the thin meter stick in our exercise, the moment of inertia depends on the position of the rotation axis:
  • **Through the center:** The formula is \( I = \frac{1}{12} M L^2 \), indicating a smaller resistance to rotation.
  • **Through one end:** The formula becomes \( I = \frac{1}{3} M L^2 \), showing greater resistance due to increased distribution of mass away from the rotation axis.
These formulas highlight how changing the rotation axis affects the moment of inertia and, consequently, the angular motion of the object.
Angular Velocity
Angular velocity measures how fast an object spins around its rotation axis. It is akin to linear velocity but applies to rotating systems. Its unit is typically radians per second (rad/s), though it can also be expressed in revolutions per minute (rev/min). In the exercise, we needed to convert the given angular velocity from revolutions per minute to radians per second. This involved multiplying the number of revolutions by \(2\pi\) to get radians and dividing the result by 60 to convert minutes into seconds.Understanding angular velocity involves visualizing rotation. Imagine a spinning top; how many turns does it make per second? That's essentially what angular velocity quantifies. A higher angular velocity means faster spinning.
Rotation Axis
The concept of a rotation axis is essential for understanding rotational dynamics. The rotation axis is an imaginary line around which an object rotates. Consider a bicycle wheel as an example. The wheel spins around its axle, which serves as its rotation axis. The position of this axis strongly influences the moment of inertia and angular momentum of the object. In the context of the meter stick problem, two scenarios were presented:
  • **Axis through the center:** Here, the stick has a balanced distribution of mass around the rotation axis, leading to a smaller moment of inertia.
  • **Axis through one end:** In this case, more mass is farther from the axis, increasing the moment of inertia.
Understanding where the rotation axis lies in relation to an object is crucial, as it affects both the dynamics and expressions used to compute rotational characteristics.

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Most popular questions from this chapter

A diver off the high board imparts an initial rotation with his body fully extended before going into a tuck and executing three back somersaults before hitting the water. If his moment of inertia before the tuck is \(16.9 \mathrm{kg} \cdot \mathrm{m}^{2}\) and after the tuck during the somersaults is \(4.2 \mathrm{kg} \cdot \mathrm{m}^{2}\) what rotation rate must he impart to his body directly off the board and before the tuck if he takes 1.4 s to execute the somersaults before hitting the water?

A satellite is spinning at 6.0 rev/s. The satellite consists of a main body in the shape of a sphere of radius \(2.0 \mathrm{m}\) and mass \(10,000 \mathrm{kg}\), and two antennas projecting out from the center of mass of the main body that can be approximated with rods of length \(3.0 \mathrm{m}\) each and mass 10 kg. The antenna's lie in the plane of rotation. What is the angular momentum of the satellite?

An Earth satellite has its apogee at \(2500 \mathrm{km}\) above the surface of Earth and perigee at \(500 \mathrm{km}\) above the surface of Earth. At apogee its speed is \(730 \mathrm{m} / \mathrm{s}\). What is its speed at perigee? Earth's radius is \(6370 \mathrm{km}\) (see below).

A solid cylinder of radius \(10.0 \mathrm{cm}\) rolls down an incline with slipping. The angle of the incline is \(30^{\circ} .\) The coefficient of kinetic friction on the surface is \(0.400 .\) What is the angular acceleration of the solid cylinder? What is the linear acceleration?

(a) What is the angular momentum of the Moon in its orbit around Earth? (b) How does this angular momentum compare with the angular momentum of the Moon on its axis? Remember that the Moon keeps one side toward Earth at all times.
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