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(II) Calculate the angular velocity \((a)\) of a clock's second hand, \((b)\) its minute hand, and \((c)\) its hour hand. State in rad/s. \((d)\) What is the angular acceleration in each case?

Short Answer

Expert verified
Second hand: 0.1047 rad/s, Minute hand: 0.001745 rad/s, Hour hand: 0.0001454 rad/s. Angular acceleration: 0 for all.

Step by step solution

01

Understanding Angular Velocity

Angular velocity is defined as the rate of change of angular displacement and is given by the formula \( \omega = \frac{\theta}{t} \), where \( \omega \) is the angular velocity, \( \theta \) the angle in radians, and \( t \) the time taken.
02

Calculating Angular Velocity of the Second Hand

The second hand completes a full circle of \( 2\pi \) radians in 60 seconds. Therefore, the angular velocity \( \omega_s \) is \( \omega_s = \frac{2\pi}{60} \approx 0.1047 \) rad/s.
03

Calculating Angular Velocity of the Minute Hand

The minute hand completes a full circle of \( 2\pi \) radians in 3600 seconds (60 minutes). Therefore, the angular velocity \( \omega_m \) is \( \omega_m = \frac{2\pi}{3600} \approx 0.001745 \) rad/s.
04

Calculating Angular Velocity of the Hour Hand

The hour hand completes a full circle of \( 2\pi \) radians in 43200 seconds (12 hours). Therefore, the angular velocity \( \omega_h \) is \( \omega_h = \frac{2\pi}{43200} \approx 0.0001454 \) rad/s.
05

Understanding Angular Acceleration

Angular acceleration is the rate of change of angular velocity with respect to time. Since the clock's hands move with a constant angular velocity, their angular acceleration is zero because their angular velocity is not changing over time.

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Key Concepts

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

Understanding Angular Acceleration
Angular acceleration describes how quickly the angular velocity of an object is changing. Think of it as the rotational equivalent of linear acceleration. For instance, if a car speeds up, its speed increases, which means it's accelerating. Similarly, angular acceleration measures how the rate of spinning speeds up or slows down over time.

The formula for angular acceleration is:
  • \[\alpha = \frac{d\omega}{dt}\]
where \(\alpha\) is the angular acceleration, \(\omega\) is the angular velocity, and \(t\) is time.

In the case of a clock's hands, they are moving at a constant speed around the clock face. This means their angular velocity is consistent and does not change. Therefore, the angular acceleration of each hand is zero, as there is no change in their angular velocity with respect to time.
Radians and Circular Measurement
Radians are a way we measure angles and are very important in the study of circular motion. Unlike degrees, which divide a circle into 360 parts, radians relate more directly to the arc length and the radius of the circle.

A full circle equals \(2\pi\) radians, which is about 6.28318. This comes from the formula:
  • Circumference of a circle = \(2\pi r\)
  • Where \(r\) is the radius of the circle
If you think about radians as chunks of the whole circle, it helps understand why a full circle is \(2\pi\) radians.

In the context of the clock, using radians helps calculate how far the hands move on the clock face as it relates to time. This is crucial for determining angular velocity.
The Function and Movement of Clocks
Clocks are an everyday example of circular motion and angular measurements. The second, minute, and hour hands each move in circular patterns, guided by the mechanisms inside the clock to provide accurate timekeeping.

Each hand works at different angular velocities:
  • The second hand completes one full revolution every 60 seconds, moving quickly and consistently.
  • The minute hand takes 3600 seconds (or one hour) for a single revolution, moving much slower.
  • The hour hand completes a circle in 43200 seconds (or 12 hours), moving even more slowly.
These movements illustrate constant angular velocity as they move at the same rate throughout their cycles.

Understanding how these hands work provides a practical example of circular motion, aiding in grasping concepts like angular velocity.
Concept of Circular Motion
Circular motion refers to the movement of an object along the circumference of a circle. This can be uniform (constant speed) or non-uniform (changing speed).

Some key terms include:
  • **Angular velocity**: the rate at which an object moves around a circle, often measured in radians per second.
  • **Angular acceleration**: measures how much faster or slower an object rotates over time.
  • **Periodic motion**: if the motion repeats at regular intervals, like the hands of a clock.
In uniform circular motion, such as that seen in clock hands, the object moves at a constant speed, meaning there's no angular acceleration. Non-uniform circular motion, like a car turning at different speeds, would involve angular acceleration.

Understanding circular motion is fundamental in physics and engineering, where it applies to everything from designing gears to analyzing celestial orbits.

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Most popular questions from this chapter

(II) A person of mass 75 kg stands at the center of a rotating merry-go-round platform of radius 3.0 m and moment of inertia \(820 kg\cdot m^2\). The platform rotates without friction with angular velocity 0.95 rad/s. The person walks radially to the edge of the platform. \((a)\) Calculate the angular velocity when the person reaches the edge. \((b)\) Calculate the rotational kinetic energy of the system of platform plus person before and after the person's walk.

(II) A nonrotating cylindrical disk of moment of inertia \(I\) is dropped onto an identical disk rotating at angular speed \(\omega \). Assuming no external torques, what is the final common angular speed of the two disks?

(II) An oxygen molecule consists of two oxygen atoms whose total mass is \(5.3 \times 10^{-26} kg\) and whose moment of inertia about an axis perpendicular to the line joining the two atoms, midway between them, is \(1.9 \times 10^{-46} kg\cdot m^2\). From these data, estimate the effective distance between the atoms.

Suppose a star the size of our Sun, but with mass 8.0 times as great, were rotating at a speed of 1.0 revolution every 9.0 days. If it were to undergo gravitational collapse to a neutron star of radius 12 km, losing \\(\frac{3}{4}\\) of its mass in the process, what would its rotation speed be? Assume the star is a uniform sphere at all times. Assume also that the thrownoff mass carries off either \((a)\) no angular momentum, or \((b)\) its proportional share \((\frac{3}{4})\) of the initial angular momentum.

(II) A 61-cm-diameter wheel accelerates uniformly about its center from 120 rpm to 280 rpm in 4.0 s. Determine \((a)\) its angular acceleration, and \((b)\) the radial and tangential components of the linear acceleration of a point on the edge of the wheel 2.0 s after it has started accelerating.

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