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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 \(\mathrm{g}\) . Take the second hand to be a slender rod rotating with constant angular velocity about one end.

Short Answer

Expert verified
The magnitude of the angular momentum is approximately \(4.712 \times 10^{-6}\, \mathrm{kg\cdot m^2/s}\).

Step by step solution

01

Understand the Problem

The exercise asks us to calculate the angular momentum of the second hand of a clock. The second hand is considered a slender rod, which rotates about the axis through the center of the clock. The length of the rod is 15.0 cm, and its mass is 6.00 g.
02

Find Angular Velocity

The angular velocity \( \omega \) of the second hand is found by recognizing that the second hand completes one full rotation (\(2\pi\) radians) in 60 seconds. Thus, \( \omega = \frac{2\pi}{60} \) rad/s.
03

Calculate the Moment of Inertia

For a slender rod rotating about one end, the moment of inertia \( I \) is \( \frac{1}{3} m L^2 \), where \( m \) is the mass in kilograms and \( L \) is the length in meters. Here, \( m = 0.006 \) kg and \( L = 0.15 \) m, so \( I = \frac{1}{3} \times 0.006 \times (0.15)^2 \).
04

Compute Angular Momentum

Angular momentum \( L \) is given by the product \( I \omega \). Now, substitute the values - moment of inertia from Step 3 and angular velocity from Step 2: \( L = \frac{1}{3} \times 0.006 \times (0.15)^2 \times \frac{2\pi}{60} \).
05

Calculate the Final Result

Perform the calculations: \( I = 0.000045 \) kg\( \cdot \)m\(^2\) and \( \omega = 0.10472 \) rad/s, giving us \( L = 0.000045 \times 0.10472 = 4.712\times10^{-6} \) kg\( \cdot \)m\(^2\)/s.

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Key Concepts

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

Slender Rod Rotation
When we discuss the rotation of a slender rod, we're talking about an object that has a much larger length compared to its width and depth. In physics, such objects are often simplified and treated as one-dimensional. The second hand of a clock is an excellent example of this, where we consider it rotating around one end. This type of motion is very characteristic of many everyday phenomena and is crucial for understanding more complex systems.
Slender rod rotation can be observed widely in mechanical and electronic devices. The simplicity of considering the rod as slender helps us apply mathematical models, which makes solving problems more straightforward. It’s important because it allows us to calculate other essential properties such as the moment of inertia, a fundamental aspect of rotational dynamics.
Moment of Inertia
In rotational dynamics, the moment of inertia is akin to mass in linear motion—it determines the resistance of an object to changes in its rotational motion. Think of it as "rotational mass." For a slender rod that rotates about one end, the formula to find the moment of inertia is \[ I = \frac{1}{3} m L^2 \]where:
  • \( m \) represents the mass of the rod, and
  • \( L \) is the length of the rod.
For instance, in our clock hand example, if we substitute the mass as 0.006 kg and length as 0.15 m into the formula, we get:\[ I = \frac{1}{3} \times 0.006 \cdot (0.15)^2 \]:showcasing how mass distribution affects rotational characteristics. The moment of inertia is crucial not just in physics exams but in understanding real-world applications like the design of rotating machinery and even amusement park rides.
Angular Velocity
When objects rotate, we describe their rotational speed in terms of angular velocity. This expresses how fast an object spins around a fixed point, and for rotations like those in clocks or wheels, it’s typically measured in radians per second. To imagine this, think about the second hand of a clock completing a full circle in 60 seconds. Understanding this, you can say it revolves through an angle of \(2\pi\) radians in that time.
The formula for angular velocity \( \omega \) is:\[ \omega = \frac{\text{angle in radians}}{\text{time in seconds}} \]Thus, \[ \omega = \frac{2\pi}{60} \approx 0.10472 \text{ rad/s} \]for a second hand. Angular velocity helps predict where an object will be after a given time and is essential for creating things like engines where timing is crucial for performance. On a deeper level, mastering angular velocity is stepping towards understanding the larger dynamics of rotating systems.

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Most popular questions from this chapter

A small block with mass 0.250 \(\mathrm{kg}\) is attached to a string passing through a hole in a frictionless, horizontal surface (see Fig. 10.48 ). The block is originally revolving in a circle with a radius of 0.800 \(\mathrm{m}\) m about the hole with a tangential speed of 4.00 \(\mathrm{m} / \mathrm{s}\) . The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 \(\mathrm{N}\) . What is the radius of the circle when the string breaks?

A \(2.00-\mathrm{kg}\) textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is \(0.150 \mathrm{m},\) to a hanging book with mass 3.00 \(\mathrm{kg}\) . The system is released from rest, and the books are observed to move 1.20 \(\mathrm{m}\) in 0.800 \(\mathrm{s}\) s. (a) What is the tension in each part of the cord? \((b)\) What is the moment of inertia of the pulley about its rotation axis?

A stiff uniform wire of mass \(M_{0}\) and length \(L_{0}\) is cut, bent, and the parts soldered together so that it forms a circular wheel having four identical spokes coming out from the center. None of the wire is wasted, and you can neglect the mass of the solder. (a) What is the moment of inertia of this wheel about an axle through its center perpendicular to the plane of the wheel? (b) If the wheel is given an initial spin with angular velocity \(\omega_{0}\) and stops uniformly in time \(T\) , what is the frictional torque at its axle?

One force acting on a machine part is \(\overrightarrow{\boldsymbol{F}}=(-5.00 \mathrm{N}) \hat{\boldsymbol{\imath}}+\) \((4.00 \mathrm{N}) \hat{\jmath}\) . The vector from the origin to the point where the force is applied is \(\vec{r}=(-0.450 \mathrm{m}) \hat{\imath}+(0.150 \mathrm{m}) \hat{\jmath} .\) (a) In a sketch, show \(\vec{r}, \vec{F},\) and the origin. (b) Use the right- hand rule to determine the direction of the torque. (c) Calculate the vector torque produced by this force. Verify that the direction of the torque is the same as you obtained in part (b).

A solid, uniform cylinder with mass 8.25 \(\mathrm{kg}\) and diameter 15.0 \(\mathrm{cm}\) is spinning at 220 \(\mathrm{rpm}\) on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and \(\mathrm{rim}\) is \(0.333 .\) What must the applied normal force be to bring the cylinder to rest after it has turned through 5.25 revolutions?

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