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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

Suppose the polar ice sheets broke free and floated toward Earth's equator without melting. What would happen to Earth's angular velocity?

A bug flying horizontally at \(1.0 \mathrm{m} / \mathrm{s}\) collides and sticks to the end of a uniform stick hanging vertically. After the impact, the stick swings out to a maximum angle of \(5.0^{\circ}\) from the vertical before rotating back. If the mass of the stick is 10 times that of the bug, calculate the length of the stick.

The precession angular velocity of a gyroscope is 1.0 rad/s. If the mass of the rotating disk is \(0.4 \mathrm{kg}\) and its radius is \(30 \mathrm{cm},\) as well as the distance from the center of mass to the pivot, what is the rotation rate in rev/s of the disk?

Three children are riding on the edge of a merry-go-round that is \(100 \mathrm{kg}\), has a 1.60 -m radius, and is spinning at \(20.0 \mathrm{rpm}\). The children have masses of \(22.0,28.0,\) and 33.0 kg. If the child who has a mass of 28.0 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm?

(a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is \(0.400 \mathrm{kg} \cdot \mathrm{m}^{2} .\) (b) He reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia if his angular velocity decreases to 1.25 rev/s. (c) Suppose instead he keeps his arms in and allows friction of the ice to slow him to 3.00 rev/s. What average torque was exerted if this takes \(15.0 \mathrm{s} ?\)
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