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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 \(\mathrm{cm}\) 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 \(4.71 \times 10^{-7}\) kg·m²/s.

Step by step solution

01

Convert Units

First, we need to convert the given units into standard metric units. Convert the length from centimeters to meters and the mass from grams to kilograms. Thus, the length \(L\) is 0.15 meters and the mass \(m\) is 0.006 kilograms.
02

Angular Velocity

The second hand completes one full revolution (360 degrees or \(2\pi\) radians) in 60 seconds. So, the angular velocity \(\omega\) is \( \frac{2\pi\, \text{rad}}{60\, \text{s}} = \frac{\pi}{30}\, \text{rad/s} \).
03

Moment of Inertia

For a slender rod rotating about one end, the moment of inertia \( I \) is given by \( I = \frac{1}{3}mL^2 \). Substitute the values: \( I = \frac{1}{3}(0.006\, \text{kg})(0.15\, \text{m})^2 = 4.5 \times 10^{-6}\, \text{kg}\cdot\text{m}^2 \).
04

Calculate Angular Momentum

The magnitude of the angular momentum \( L \) can be calculated using \( L = I\omega \). Substitute the values: \( L = (4.5 \times 10^{-6}\, \text{kg}\cdot\text{m}^2) \times \left( \frac{\pi}{30}\, \text{rad/s} \right) = 4.71 \times 10^{-7}\, \text{kg}\cdot\text{m}^2/\text{s} \).

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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, often denoted by the symbol \( I \), is a measure of an object's resistance to changes in its rotational motion. It is analogous to mass in linear motion, but instead of mass, the moment of inertia takes into account how mass is distributed relative to the axis of rotation. In essence, it tells us how difficult it is to change the angular velocity of a rotating object.

For different shapes and axis points, the moment of inertia is calculated using different formulas. For instance, for a slender rod rotating about one end, like the second hand of a clock, the formula used is:

\[ I = \frac{1}{3} mL^2 \]

where \( m \) is the mass of the object and \( L \) is its length. The structure of the object and the axis of rotation impact the value of \( I \). Hence, understanding how to compute the moment of inertia helps in predicting the behavior of rotating systems.

When dealing with real-life problems, always ensure to convert units to the standard metric units (meters and kilograms in this case) to maintain consistency in calculations.
Angular Velocity
Angular velocity is a crucial concept in understanding rotational motion. It measures how fast an object rotates or revolves relative to another point, often concentrating on how quickly an angle is changing in a rotational system. Angular velocity, usually represented by \( \omega \), tells us the rate at which an object turns through an angle over time. It's typically expressed in radians per second.

To illustrate, consider the second hand of a clock. It completes a full rotation, which is \( 2\pi \) radians, once every 60 seconds. This means its angular velocity is:

\[ \omega = \frac{2\pi}{60} = \frac{\pi}{30} \text{ rad/s} \]

This concept helps us analyze how quickly objects are rotating, and it plays a critical role alongside moment of inertia in determining the angular momentum of rotating objects. The subtle relationship between these parameters provides deeper insights into the dynamics of any rotating system.

Understanding angular velocity opens the door to analyzing situations where rotational symmetry is involved, adding another layer to studying motions.
Rotational Motion
Rotational motion is a key topic in physics, dealing with objects that rotate about an axis. This type of motion is everywhere in our daily lives, from the earth's rotation to wheels turning. Grasping rotational motion involves understanding how objects turn, speeds of rotation, and the effects of applied forces over time.

Core principles in rotational motion include not just angular velocity and moment of inertia, but also torque, which is the force that causes an object to rotate. When an object is rotating, it's not just about how fast it's spinning, but also understanding how its mass is distributed, which directly impacts its rotational characteristics.

Using our clock example again, each component (moment of inertia and angular velocity) sits as a part of the bigger picture of rotational dynamics. With rotational motion, one must consider the balance of forces and how they apply around a pivot or axis.
  • Key terms: angular displacement, which is the change angle as an object rotates.
  • Angular acceleration, the rate of change of angular velocity.
  • These lie at the heart of analyzing and predicting sophisticated movements in any rotating system.
By learning these, you're well-equipped to tackle even more complex rotational systems, making these principles foundational to the study of physics and engineering.

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Most popular questions from this chapter

A lawn roller in the form of a thin-walled, hollow cylinder with mass \(M\) is pulled horizontally with a constant horizontal force \(F\) applied by a handle attached to the axle. If it rolls without slipping, find the acceleration and the friction force.

A \(2.20-\) kg hoop 1.20 \(\mathrm{m}\) in diameter is rolling to the right without slipping on a horizontal floor at a steady 3.00 \(\mathrm{rad} / \mathrm{s}\) . (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 Gyroscope on the Moon. A certain gyroscope precesses at a rate of 0.50 \(\mathrm{rad} / \mathrm{s}\) when used on earth. If it were taken to a lunar base, where the acceleration due to gravity is \(0.165 g,\) what would be its precession rate?

A small 10.0-g bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass 50.0 \(\mathrm{g}\) and is 100 \(\mathrm{cm}\) in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of 20.0 \(\mathrm{cm} / \mathrm{s}\) relative to the table. (a) What is the angular speed of the bar just after the frisky insect leaps? (b) What is the total kinetic energy of the system just after the bug leaps? (c) Where does this energy come from?

A 15.0 -kg bucket of water is suspended by a very light rope wrapped around a solid uniform cylinder 0.300 \(\mathrm{m}\) in diameter with mass 12.0 \(\mathrm{kg} .\) The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 10.0 \(\mathrm{m}\) to the water. (a) What is the tension in the rope while the bucket is falling? (b) With what speed does the bucket strike the water? (c) What is the time of fall? (d) While the bucket is falling, what is the force exerted on the cylinder by the axle?

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