/*! 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 37 Find the magnitude of the angula... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of the clock face. The clock hand has a length of 15.0 cm and a mass of 6.00 g. Take the second hand to be a slender rod rotating with constant angular velocity about one end.

Short Answer

Expert verified
The angular momentum is approximately \( 4.71 \times 10^{-6} \text{ kg·m}^2/\text{s} \).

Step by step solution

01

Understand the Problem

The problem requires us to find the magnitude of the angular momentum for the second hand considered as a slender rod rotating about an axis through the center of the clock. We need to use the properties of the second hand: length of 15 cm and mass of 6 g.
02

Convert Units

The length of the second hand is given in cm, and the mass is given in grams. We need to convert these into meters and kilograms respectively. - Length: 15 cm = 0.15 m - Mass: 6 g = 0.006 kg
03

Calculate Moment of Inertia

For a slender rod of length \( L \) and mass \( m \) rotating about one end, the moment of inertia \( I \) is given by:\[ I = \frac{1}{3} m L^2 \] Substituting the given values:\[ I = \frac{1}{3} \times 0.006 \text{ kg} \times (0.15 \text{ m})^2 = 4.5 \times 10^{-5} \text{ kg·m}^2 \]
04

Calculate Angular Velocity

The second hand completes a full circle in 60 seconds, giving it an angular velocity \( \omega \) of:\[ \omega = \frac{2\pi \text{ radians}}{60 \text{ seconds}} = \frac{\pi}{30} \text{ rad/s} \]
05

Calculate Angular Momentum

The angular momentum \( L \) of the second hand can be calculated using:\[ L = I \cdot \omega \]Substituting the values from previous steps:\[ L = 4.5 \times 10^{-5} \text{ kg·m}^2 \times \frac{\pi}{30} \text{ rad/s} = 4.71 \times 10^{-6} \text{ kg·m}^2/\text{s} \]
06

Solution Verification

We have calculated the angular momentum using the formulas for moment of inertia of a rod and the angular velocity of the second hand correctly. Thus, the computed value should be reliable.

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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 crucial concept in understanding rotational dynamics. It represents the rotational counterpart to mass in linear motion. Essentially, it's a measure of an object's resistance to changes in its rotation.
  • The moment of inertia depends on the distribution of mass in an object and the axis about which it rotates.
  • For different shapes and axes, there are different formulas to calculate the moment of inertia. In this exercise, the formula for a slender rod rotating about one end is used.
For a slender rod with mass \( m \) and length \( L \), the moment of inertia \( I \) about one end is calculated as:\[ I = \frac{1}{3} m L^2 \]This formula indicates that both the mass and length of the rod greatly influence its resistance to rotational changes. The longer or heavier the rod, the larger the moment of inertia, meaning it will be harder to spin or stop it from spinning once it starts.
Angular Velocity
Angular velocity describes the speed of rotation of an object around a certain axis. It is the rate at which the angular position or orientation of an object changes with time. The unit of angular velocity is expressed as radians per second (rad/s).
  • In the context of this exercise, the second hand of a clock rotates at a consistent speed.
  • It completes one full revolution every 60 seconds, reflecting a uniform circular motion.
The angular velocity \( \omega \) can be derived from the formula:\[ \omega = \frac{2\pi}{T} \]where \( T \) is the period of rotation in seconds. For the second hand:\[ \omega = \frac{2\pi}{60} = \frac{\pi}{30} \text{ rad/s} \] This consistent angular velocity simplifies calculations in rotational dynamics, making it a foundational measure in understanding rotational motion behaviors.
Rotational Dynamics
Rotational dynamics is the area of physics examining the effects of torques and angular motion of objects. It extends principles of linear dynamics to rotating objects.
  • Key factors include torque, angular momentum, and moments of inertia.
  • Rotational dynamics help predict how forces applied to objects will affect their spinning motion.
In this specific exercise, the angular momentum \( L \) is important because it combines the moment of inertia and angular velocity to describe the overall motion:\[ L = I \cdot \omega \]Angular momentum provides insight into the conservation law, where the total angular momentum in a closed system remains constant unless acted upon by an external torque. This principle is a powerful tool for analyzing both simple and complex systems in rotational motion. Understanding these principles enables us to predict and manipulate rotational behaviors in practical applications like mechanical engineering and celestial motions.

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Most popular questions from this chapter

A yo-yo is made from two uniform disks, each with mass \(m\) and radius \(R\), connected by a light axle of radius \(b\). A light, thin string is wound several times around the axle and then held stationary while the yo-yo is released from rest, dropping as the string unwinds. Find the linear acceleration and angular acceleration of the yo-yo and the tension in the string.

A 2.20-kg hoop 1.20 m in diameter is rolling to the right without slipping on a horizontal floor at a steady 2.60 rad/s. (a) How fast is its center moving? (b) What is the total kinetic energy of the hoop? (c) Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom. (d) Find the velocity vector for each of the points in part (c), but this time as viewed by someone moving along with the same velocity as the hoop.

A 392-N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 25.0 rad/s. The radius of the wheel is 0.600 m, and its moment of inertia about its rotation axis is 0.800MR\(^2\). Friction does work on the wheel as it rolls up the hill to a stop, a height \(h\) above the bottom of the hill; this work has absolute value 2600 J. Calculate \(h\).

You are designing a slide for a water park. In a sitting position, park guests slide a vertical distance \(h\) down the waterslide, which has negligible friction. When they reach the bottom of the slide, they grab a handle at the bottom end of a 6.00-m-long uniform pole. The pole hangs vertically, initially at rest. The upper end of the pole is pivoted about a stationary, frictionless axle. The pole with a person hanging on the end swings up through an angle of 72.0\(^\circ\), and then the person lets go of the pole and drops into a pool of water. Treat the person as a point mass. The pole's moment of inertia is given by \(I = \frac{1}{3} ML^2\), where \(L =\) 6.00 m is the length of the pole and \(M =\) 24.0 kg is its mass. For a person of mass 70.0 kg, what must be the height h in order for the pole to have a maximum angle of swing of 72.0\(^\circ\) after the collision?

A cord is wrapped around the rim of a solid uniform wheel 0.250 m in radius and of mass 9.20 kg. A steady horizontal pull of 40.0 N to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) Compute the angular acceleration of the wheel and the acceleration of the part of the cord that has already been pulled off the wheel. (b) Find the magnitude and direction of the force that the axle exerts on the wheel. (c) Which of the answers in parts (a) and (b) would change if the pull were upward instead of horizontal?

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