/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 M A particle of mass \(0.400 \ma... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 66

M A particle of mass \(0.400 \mathrm{~kg}\) is attached to the \(100-\mathrm{cm}\) mark of a meter stick of mass \(0.100 \mathrm{~kg}\). The meter stick rotates on a horizontal, frictionless table with an angular speed of \(4.00 \mathrm{rad} / \mathrm{s}\). Calculate the angular momentum of the system when the stick is pivoted about an axis (a) perpendicular to the table through the \(50.0-\mathrm{cm}\) mark and (b) perpendicular to the table through the \(0-\mathrm{cm}\) mark.

Short Answer

Expert verified
The angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the \(50.0-\mathrm{cm}\) mark is \(0.48332 \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s}\). When pivoted at the \(0-\mathrm{cm}\) mark, it is \(1.73332 \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s}\).

Step by step solution

01

Determine the moment of inertia

The moment of inertia of the meter stick about the pivot point A, the \(50.0-\mathrm{cm}\) mark, can be represented as \(I_{stick,A} = \frac{1}{12} m_{stick} L^2 + m_{stick} d^2\), which equals to \(\frac{1}{12} (0.1kg) (1m)^2 + (0.1kg)(0.5m)^2 = 0.02083 \mathrm{kg} \cdot \mathrm{m}^2\). m_stick is the mass of the stick, L is the length of the stick and d is the distance from the pivot point to the center of mass of the stick. The moment of inertia of the particle about point A equals to \(I_{particle,A}= m_{particle} r^2\), where m_particle is the mass of the particle and r is the distance between the pivot point and the particle, thus it equals to \(0.4kg \cdot (0.5m)^2 = 0.1 \mathrm{kg} \cdot \mathrm{m}^2\). Therefore, the total moment of inertia about point A is the sum of the both, \(I_{total,A} = I_{stick,A} + I_{particle,A} = 0.02083 \mathrm{kg} \cdot \mathrm{m}^2 + 0.1 \mathrm{kg} \cdot \mathrm{m}^2 = 0.12083 \mathrm{kg} \cdot \mathrm{m}^2\).
02

Compute the angular momentum about point A

Next, calculate the angular momentum for the system about point A by multiplying the total moment of inertia with the angular speed according to the relation \(L=\omega I\). As a result, \(L_A = \omega \cdot I_{total,A} = 4.00 \mathrm{rad/s} \cdot 0.12083 \mathrm{kg} \cdot \mathrm{m}^2 = 0.48332 \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s}\).
03

Find the moment of inertia for point B

Now, repeat the same process in steps 1 and 2 to determine the angular momentum about point B, the \(0-\mathrm{cm}\) mark. The moment of inertia of the stick around this point is \(I_{stick,B} = \frac{1}{3} m_{stick} L^2 = \frac{1}{3}(0.1kg)(1m)^2 = 0.03333 \mathrm{kg} \cdot \mathrm{m}^2\). The moment of inertia of the particle about point B is, \(I_{particle,B} = m_{particle} R^2 = 0.4kg \cdot (1m)^2 = 0.4 \mathrm{kg} \cdot \mathrm{m}^2\). Thus, the total moment of inertia about point B is \(I_{total,B} = I_{stick,B} + I_{particle,B} = 0.03333 \mathrm{kg} \cdot \mathrm{m}^2 + 0.4 \mathrm{kg} \cdot \mathrm{m}^2 = 0.43333 \mathrm{kg} \cdot \mathrm{m}^2\).
04

Calculate the angular momentum about point B

Finally, the angular momentum about point B can be found out, \(L_B = \omega \cdot I_{total,B} = 4.00 \mathrm{rad/s} \cdot 0.43333 \mathrm{kg} \cdot \mathrm{m}^2 = 1.73332 \mathrm{kg} \cdot \mathrm{m}^2/\mathrm{s}\).

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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, represented by the symbol 'I', is a critical value in the field of rotational motion. It quantifies an object's resistance to changes in its rotational speed. Think of it as the rotational equivalent of mass in linear motion. The higher the moment of inertia, the harder it is to start or stop the rotation of an object.

In the original exercise, we calculate the moment of inertia of both a meter stick and a particle attached to it. The meter stick鈥檚 moment of inertia depends on its mass and the distribution of this mass relative to the pivot point. The formula for a rod rotating about its center is \( I = \frac{1}{12} m L^2 \) and when it's rotating about its end, it changes to \( I = \frac{1}{3} m L^2 \). The particle's moment of inertia is calculated using \( I_{particle} = m r^2 \), where 'm' is the mass and 'r' is the distance from the pivot.

To improve understanding, it鈥檚 important to remember that for any object or system, the overall moment of inertia is the sum of the individual moments of all mass elements making up the system.
Angular Speed
Angular speed is a measure of how quickly an object rotates. Expressed in radians per second (rad/s), it describes the angle through which an object spins in a unit of time. For a given angular speed, parts of an object farther from the axis of rotation move faster than parts closer to the axis.

In our exercise, the angular speed given is \(4.00 \mathrm{rad/s}\). Applying this to the entire system, consisting of the meter stick and particle, allows us to calculate the angular momentum. This uniform angular speed across the system simplifies the calculation. To gain a better grasp on this, visualize a spinning top; it spins around its axis at a constant rate鈥攖his rate is its angular speed.
Rotational Motion
Rotational motion is the circular movement of an object around a center or axis. In this kind of motion, all points in the object follow circular paths with the axis of rotation being stationary. Rotational motion is present in many physical situations, from merry-go-rounds to the rotation of celestial bodies.

The exercise discusses a meter stick pivoting around two different axis points. Each point creates a different scenario for angular motion, illustrated by the different moments of inertia calculated. When we combine the concepts of moment of inertia and angular speed, we obtain the angular momentum. Angular momentum is a fundamental aspect of rotational motion, representing the quantity of rotation of a body and is conserved in the absence of external torques. For angular momentum, recall the equation \( L = I \omega \), which we used to solve the problem. Encouraging students to visualize these scenarios can significantly aid in their understanding of rotational dynamics.

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Most popular questions from this chapter

A solid, uniform disk of radius \(0.250 \mathrm{~m}\) and mass \(55.0 \mathrm{~kg}\) rolls down a ramp of length \(4.50 \mathrm{~m}\) that makes an angle of \(15.0^{\circ}\) with the horizontal. The disk starts from rest from the top of the ramp. Find (a) the speed of the disk's center of mass when it reaches the bottom of the ramp and (b) the angular speed of the disk at the bottom of the ramp.

An object of mass \(M=\) \(12.0 \mathrm{~kg}\) is attached to a cond that is wrapped around a wheel of radius \(r=10.0 \mathrm{~cm}\) (Fig. P8.70). The acceleration of the object down the frictionless incline is measured to be \(a=2.00 \mathrm{~m} / \mathrm{s}^{2}\) andthe incline makes an angle \(\theta=37.0^{\circ}\) with the horizontal. Assuming the axle of the wheel to be frictionless, determine (a) the tension in the rope, (b) the moment of inertia of the wheel, and (c) the angular speed of the wheel \(2.00 \mathrm{~s}\) after it begins rotating, starting from rest.

S A uniform solid cylinder of mass \(M\) and radius \(R\) rotates on a frictionless horizontal axle (Fig. P8.81). Two objects with equal masses \(m\) hang from light cords wrapped around the cylinder. If the system is released from rest, find (a) the tension in eachcord and (b) the acceleration of each object after the objects have descended a distance \(h\).

QIC I S A painter climbs a ladder leaning against a smooth wall. At a certain height, the ladder is on the verge of slipping. (a) Explain why the force exerted bythe vertical wall on the ladder is horizontal. (b) If the ladder of length \(L\) leans at an angle \(\theta\) with the horizontal, what is the lever arm for this horizontal force with the axis of rotation taken at the base of the ladder? (c) If the ladder is uniform, what is the lever arm for the force of gravity acting on the ladder? (d) Let the mass of the painter be \(80 \mathrm{~kg}, L=4.0 \mathrm{~m}\), the ladder's mass be \(30 \mathrm{~kg}, \theta=53^{\circ}\), and the coefficient of friction between ground and ladder be \(0.45\). Find the maximum distance the painter can climb up the ladder.

A constant torque of \(25.0 \mathrm{~N} \cdot \mathrm{m}\) is applied to a grindstone whose moment of inertia is \(0.130 \mathrm{~kg} \cdot \mathrm{m}^{2}\). Using energy principles and neglecting friction, find the angular speed after the grindstone has made \(15.0\) revolutions. Hint: The angular equivalent of \(W_{\text {net }}=F \Delta x=\) \(\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}\) is \(W_{\text {net }}=\tau \Delta \theta=\frac{1}{2} I \omega_{f}^{2}-\frac{1}{2} I \omega_{i}^{2}\). You should convince yourself that this last relationship is correct.
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