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A bicycle is moving with a speed of \(4.02 \mathrm{~m} / \mathrm{s}\). If the radius of the front wheel is \(0.450 \mathrm{~m}\), how long does it take for that wheel to make a complete revolution? a) \(0.703 \mathrm{~s}\) b) \(1.23 \mathrm{~s}\) c) \(2.34 \mathrm{~s}\) d) \(4.04 \mathrm{~s}\) e) \(6.78 \mathrm{~s}\)

Short Answer

Expert verified
Answer: a) The time taken for one complete revolution of the front wheel is approximately 0.703 seconds.

Step by step solution

01

Find the bicycle's speed

Given the bicycle's speed is \(4.02 \mathrm{~m} / \mathrm{s}\). This means that the outer edge of the front wheel also moves at this speed.
02

Find the radius of the front wheel

The radius of the front wheel is given as \(0.450 \mathrm{~m}\).
03

Calculate the circumference of the front wheel

To find the circumference (distance covered in one complete revolution), we need to use the formula \(C=2\pi r\) where \(C\) is the circumference, \(\pi\) is a constant approximately equal to \(3.1416\), and \(r\) is the radius. In this case, the radius is \(0.450 \mathrm{~m}\). \(C=2\pi r = 2 \times 3.1416 \times 0.450 \mathrm{~m} \approx 2.83 \mathrm{~m}\).
04

Determine the time taken for one complete revolution

Now that we have the circumference and the speed, we can find the time taken for the wheel to make one complete revolution using the formula: \(time = \frac{distance}{speed}\), where \(distance\) is the circumference of the wheel and \(speed\) is the bicycle's speed. Substituting the values, we get: \(time = \frac{2.83 \mathrm{~m}}{4.02 \mathrm{~m} / \mathrm{s}} \approx 0.703 \mathrm{~s}\). So the correct answer is a) \(0.703 \mathrm{~s}\).

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Key Concepts

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

Circumference of a Circle
The circumference of a circle is one of the fundamental concepts in geometry and can be often found in physics problems related to circular motion. It refers to the distance around the edge of the circle. To calculate it, we use the formula:

\[C = 2\text{\textpi}r\],

where \(C\) represents the circumference, \(\text{\textpi}\) is Pi - a mathematical constant approximately equal to 3.1416, and \(r\) is the radius of the circle, which is the distance from the center to any point on its edge.
In our bicycle wheel example, with the radius of 0.450 meters, we apply the formula to find out how far the wheel travels in one full rotation. It's important to understand that this circumference is also the distance that the bicycle travels forward with every full spin of the wheel. By learning to connect the concept of circumference with physical movement, students can better grasp how geometry is used to describe motion in our day to day life.
Angular Velocity
Angular velocity is a measure of how fast an object rotates or revolves relative to another point, which we say is the angular speed of the object. In the context of uniform circular motion, angular velocity is typically expressed in radians per second (rad/s). It is given by the formula:

\[ \text{\textomega} = \frac{\theta}{t} \],

where \(\text{\textomega}\) represents the angular velocity, \(\theta\) is the angle in radians through which a point or line has been rotated in a specific time duration \(t\). Remember that one complete revolution is \(2\text{\textpi}\) radians.
For example, if our bicycle wheel makes one complete revolution, it rotates through an angle of \(2\text{\textpi}\) radians. If we know the time taken to complete this revolution from our problem, we can calculate the angular velocity. This term is crucial for understanding rotational dynamics in physics and is also applied widely in engineering concepts such as motor speeds, and wheel rotations.
Uniform Circular Motion
Uniform circular motion describes an object moving in a circular path with constant speed. The velocity of an object in uniform circular motion is constantly changing due to its direction, but the speed (the magnitude of velocity) is constant. This concept is vital in understanding various physical phenomena and in designing mechanical systems that undergo circular paths.

Objects in uniform circular motion experience an inward force called centripetal force that keeps them moving in the circle. This force is always perpendicular to the displacement of the object and thus does not do work—meaning it does not change the kinetic energy of the object, and consequently, the speed remains constant. Understanding this helps students see why the speed – the scalar quantity – is constant, but the velocity – a vector quantity which includes direction – is always changing due to the object's continuous change in direction.
The next time you see an object like the wheel of a bicycle traveling in its path, try to visualize the principles of uniform circular motion in action. Comprehending the principle of uniform circular motion lays the foundation not just for solving physics problems but also in appreciating the forces and motion in the physical world around us.

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Most popular questions from this chapter

A wagon wheel is made entirely of wood. Its components consist of a rim, 12 spokes, and a hub. The rim has mass \(5.20 \mathrm{~kg}\), outer radius \(0.900 \mathrm{~m}\), and inner radius \(0.860 \mathrm{~m}\). The hub is a solid cylinder with mass \(3.40 \mathrm{~kg}\) and radius \(0.120 \mathrm{~m}\). The spokes are thin rods of mass \(1.10 \mathrm{~kg}\) that extend from the hub to the inner side of the rim. Determine the constant \(c=I / M R^{2}\) for this wagon wheel.

A light rope passes over a light, frictionless pulley. One end is fastened to a bunch of bananas of mass \(M\), and a monkey of the same mass clings to the other end. The monkey climbs the rope in an attempt to reach the bananas. The radius of the pulley is \(R\). a) Treating the monkey, bananas, rope, and pulley as a system, evaluate the net torque about the pulley axis. b) Using the result of part (a), determine the total angular momentum about the pulley axis as a function of time.

A space station is to provide artificial gravity to support long-term habitation by astronauts and cosmonauts. It is designed as a large wheel, with all the compartments in the rim, which is to rotate at a speed that will provide an acceleration similar to that of terrestrial gravity for the astronauts (their feet will be on the inside of the outer wall of the space station and their heads will be pointing toward the hub). After the space station is assembled in orbit, its rotation will be started by the firing of a rocket motor fixed to the outer rim, which fires tangentially to the rim. The radius of the space station is \(R=50.0 \mathrm{~m}\) and the mass is \(M=2.40 \cdot 10^{5} \mathrm{~kg} .\) If the thrust of the rocket motor is \(F=1.40 \cdot 10^{2} \mathrm{~N},\) how long should the motor fire?

It is sometimes said that if the entire population of China stood on chairs and jumped off simultaneously, it would alter the rotation of the Earth. Fortunately, physics gives us the tools to investigate such speculations. a) Calculate the moment of inertia of the Earth about its axis. For simplicity, treat the Earth as a uniform sphere of mass \(m_{\mathrm{E}}=5.977 \cdot 10^{24} \mathrm{~kg}\) and radius \(6371 \mathrm{~km}\). b) Calculate an upper limit for the contribution by the population of China to the Earth's moment of inertia, by assuming that the whole group is at the Equator. Take the population of China to be 1.30 billion people, of average mass \(70.0 \mathrm{~kg}\) c) Calculate the change in the contribution in part (b) associated with a \(1.00-\mathrm{m}\) simultaneous change in the radial position of the entire group. d) Determine the fractional change in the length of the day the change in part (c) would produce.

A projectile of mass \(m\) is launched from the origin at speed \(v_{0}\) and an angle \(\theta_{0}\) above the horizontal. Air resistance is negligible. a) Calculate the angular momentum of the projectile about the origin. b) Calculate the rate of change of this angular momentum. c) Calculate the torque acting on the projectile, about the origin, during its flight.

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