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Chapter 9: Problem 67

While riding a multispeed bicycle, the rider can select the radius of the rear sprocket that is fixed to the rear axle. The front sprocket of a bicycle has radius \(12.0 \mathrm{~cm} .\) If the angular speed of the front sprocket is 0.600 rev \(/ \mathrm{s},\) what is the radius of the rear sprocket for which the tangential speed of a point on the rim of the rear wheel will be \(5.00 \mathrm{~m} / \mathrm{s} ?\) The rear wheel has radius \(0.330 \mathrm{~m}\).

Short Answer

Expert verified
To find the radius of the rear sprocket, substitute the given values into the derived formula i.e., \(R_3 = (R_1 \cdot \omega_1 \cdot R_2) / v_2\). Therefore, solve for \(R_3\).

Step by step solution

01

Analyzing given quantities

Let's denote the following: \(R_1\) as front sprocket radius (12.0 cm = 0.12 m), \(\omega_1\) as angular speed of the front sprocket (0.600 rev/s), \(v_2\) as desired tangential speed (5.00 m/s), \(R_2\) as rear wheel radius (0.330 m), \(R_3\) as rear sprocket radius. Note that you need to convert the angular speed from rev/s to rad/s by multiplying it by \(2\pi\) (since 1 revolution is \(2\pi\) radians). So, \(\omega_1 = 0.600 \times 2\pi\) rad/s.
02

Applying Tangential Speed Formula

Since the chain connects the front sprocket and the rear sprocket, the tangential speed at the rim of both sprockets should be equal. This gives us the equation: \(v_1 = v_3\), where \(v_1 = R_1 \cdot \omega_1\) (tangential speed at the rim of the front sprocket), \(v_3 = R_3 \cdot \omega_2\) (tangential speed at the rim of the rear sprocket). It's given that the tangential speed at the rim of the rear wheel is equal to \(v_2\) = 5.00 m/s. So, it also equals \(v_3\), hence \(R_3 \cdot \omega_2 = v_2\).
03

Solving for Rear Sprocket Radius

First, calculate \(v_1\) from the formula in step 2. Then set \(v_1 = v_2\) and solve for \(R_3\). This gives us the formula: \(R_3 = v_2 / \omega_2\). Since \(v_2\) = \(v_1\), we have \(R_3 = (R_1 \cdot \omega_1) / \omega_2\). Lastly, note that the angular speed of the rear sprocket (\(\omega_2\)) is the same as the angular speed of the rear wheel, which equals \(v_2 / R_2\). Substituting this into the equation gives us the radius of the rear sprocket: \(R_3 = (R_1 \cdot \omega_1 \cdot R_2) / v_2\).
04

Substituting Numerical values

Substitute the given values: \(R_1 = 0.12\) m, \(\omega_1 = 0.600 \times 2\pi\) rad/s, \(R_2 = 0.330\) m, and \(v_2 = 5.00\) m/s into the formula to find \(R_3\).

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Key Concepts

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

Angular Speed
Angular speed is a measure of how fast an object rotates or revolves relative to another point. It is common in rotational dynamics and is represented by the symbol \( \omega \). Angular speed is usually measured in radians per second (rad/s). To convert from revolutions per second (rev/s) to radians per second, we multiply by \(2\pi\), since one full revolution is \(2\pi\) radians.
For example, if a bicycle front sprocket rotates at 0.600 rev/s, the angular speed becomes \( \omega_1 = 0.600 \times 2\pi \) rad/s. Angular speed helps in calculating other relevant rotational dynamics parameters, like tangential speed.
**Key Points:**
  • Describes how fast an object spins around an axis.
  • Units: radians per second (rad/s).
  • Conversion: 1 revolution = \(2\pi\) radians.
Tangential Speed
Tangential speed is the linear speed of a point situated on a rotating object and is perpendicular to the radius at any point. It relates to how fast a point on the edge of the object is moving through space. The formula for tangential speed \( v \) is \( v = r \cdot \omega \), where \( r \) is the radius and \( \omega \) is the angular speed of rotation.

When dealing with bicycles, tangential speed is crucial. It helps determine the speed at which the bike tires move the bicycle along a surface, impacting overall velocity. It maintains the same value across the chain for both front and rear sprockets. This unusual characteristic stems from the fact that the chain links travel across both sprockets without slipping.
**Important Features:**
  • Refers to the speed of a point along the edge of a circular path.
  • Calculated using the radius and angular speed (\( v = r \cdot \omega \)).
  • Essential for understanding bicycle motion, as it describes how wheel rotation translates into linear bike movement.
Bicycle Mechanics
Bicycle mechanics involves understanding how various parts of a bicycle work together to create smooth and efficient motion. A key component is the sprocket system which consists of the front and rear sprockets interconnected via a chain. Altering the size of these sprockets changes the mechanical advantage, which in turn affects speed and force.

In multispeed bicycles, the rear sprocket can be swapped to change the gear ratio. This adjustment allows the cyclist to either pedal easier (for uphill riding) or harder (for speed), shifting how power is transmitted from the rider to the wheels.
**Core Aspects:**
  • Sprockets and chains are integral in transferring power from the pedals to the wheels.
  • Changing sprocket size alters the bike's gear ratio, impacting speed and ease of pedaling.
  • Effective cycling requires an understanding of these adjustments for different terrains and riding conditions.
Sprocket Radius Calculation
Calculating the radius of the rear sprocket in a bicycle is essential for tuning the bike's performance to match desired riding conditions. Given a known tangential speed and rear wheel radius, you can determine the appropriate rear sprocket radius to achieve that speed. The relationship involves using the formula \( R_3 = \frac{(R_1 \cdot \omega_1 \cdot R_2)}{v_2} \), where \( R_1 \) is the radius of the front sprocket, and \( \omega_1 \) is its angular speed.

Knowing this formula helps you balance the bike's speed versus effort required to pedal, optimizing the ride for conditions like flat roads or steep hills.
**Reasons to Calculate:**
  • Determine the gear ratio necessary for specific speed targets.
  • Optimize pedaling efficiency by selecting the correct sprocket size.
  • Balance between speed and pedaling effort for different cycling needs.

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Most popular questions from this chapter

\(9.88^{\circ}\) DATA You are analyzing the motion of a large flywheel that has radius \(0.800 \mathrm{~m}\). In one test run, the wheel starts from rest and turns in a horizontal plane with constant angular acceleration. An accelerometer on the rim of the flywheel measures the magnitude of the resultant acceleration \(a\) of a point on the rim of the flywheel as a function of the angle \(\theta-\theta_{0}\) through which the wheel has turned. You collect these results: $$ \begin{array}{l|cccccccc} \boldsymbol{\theta}-\boldsymbol{\theta}_{\mathbf{0}}(\mathbf{r a d}) & 0.50 & 1.00 & 1.50 & 2.00 & 2.50 & 3.00 & 3.50 & 4.00 \\ \hline \boldsymbol{a}\left(\mathbf{m} / \mathbf{s}^{\mathbf{2}}\right) & 0.678 & 1.07 & 1.52 & 1.98 & 2.45 & 2.92 & 3.39 & 3.87 \end{array} $$ Construct a graph of \(a^{2}\left(\right.\) in \(\left.\mathrm{m}^{2} / \mathrm{s}^{4}\right)\) versus \(\left(\theta-\theta_{0}\right)^{2}\) (in rad \(^{2}\) ). (a) What are the slope and \(y\) -intercept of the straight line that gives the best fit to the data? (b) Use the slope from part (a) to find the angular acceleration of the flywheel. (c) What is the linear speed of a point on the rim of the flywheel when the wheel has turned through an angle of \(135^{\circ} ?\) (d) When the flywheel has turned through an angle of \(90.0^{\circ},\) what is the angle between the linear velocity of a point on its rim and the resultant acceleration of that point?

On a compact disc (CD), music is coded in a pattern of tiny pits arranged in a track that spirals outward toward the rim of the disc. As the disc spins inside a CD player, the track is scanned at a constant linear speed of \(v=1.25 \mathrm{~m} / \mathrm{s} .\) Because the radius of the track varies as it spirals outward, the angular speed of the disc must change as the \(\mathrm{CD}\) is played. (See Exercise \(9.20 .\) ) Let's see what angular acceleration is required to keep \(v\) constant. The equation of a spiral is \(r(\theta)=r_{0}+\beta \theta,\) where \(r_{0}\) is the radius of the spiral at \(\theta=0\) and \(\beta\) is a constant. On a \(\mathrm{CD}, r_{0}\) is the inner radius of the spiral track. If we take the rotation direction of the CD to be positive, \(\beta\) must be positive so that \(r\) increases as the disc turns and \(\theta\) increases. (a) When the disc rotates through a small angle \(d \theta,\) the distance scanned along the track is \(d s=r d \theta .\) Using the above expression for \(r(\theta),\) integrate \(d s\) to find the total distance \(s\) scanned along the track as a function of the total angle \(\theta\) through which the disc has rotated. (b) since the track is scanned at a constant linear speed \(v,\) the distance \(s\) found in part (a) is equal to vi. Use this to find \(\theta\) as a function of time. There will be two solutions for \(\theta ;\) choose the positive one, and explain why this is the solution to choose. (c) Use your expression for \(\theta(t)\) to find the angular velocity \(\omega_{z}\) and the angular acceleration \(\alpha_{z}\) as functions of time. Is \(\alpha_{z}\) constant? (d) On a CD, the inner radius of the track is \(25.0 \mathrm{~mm}\), the track radius increases by \(1.55 \mu \mathrm{m}\) per revolution, and the playing time is \(74.0 \mathrm{~min} .\) Find \(r_{0}, \beta,\) and the total number of revolutions made during the playing time. (e) Using your results from parts (c) and (d), make graphs of \(\omega_{z}\) (in rad/s) versus \(t\) and \(\alpha_{z}\) (in rad/s \(^{2}\) ) versus \(t\) between \(t=0\) and \(t=74.0 \mathrm{~min}\)

A disk of radius \(25.0 \mathrm{~cm}\) is free to turn about an axle perpendicular to it through its center. It has very thin but strong string wrapped around its rim, and the string is attached to a ball that is pulled tangentially away from the rim of the disk (Fig. \(\mathbf{P 9 . 6 1}\) ). The pull increases in magnitude and produces an acceleration of the ball that obeys the equation \(a(t)=A t,\) where \(t\) is in seconds and \(A\) is a constant. The cylinder starts from rest, and at the end of the third second, the ball's acceleration is \(1.80 \mathrm{~m} / \mathrm{s}^{2}\). (a) Find \(A\). (b) Express the angular acceleration of the disk as a function of time. (c) How much time after the disk has begun to turn does it reach an angular speed of \(15.0 \mathrm{rad} / \mathrm{s} ?\) (d) Through what angle has the disk turned just as it reaches \(15.0 \mathrm{rad} / \mathrm{s} ?\) (Hint: See Section \(2.6 .)\)

A rotating flywheel has moment of inertia \(12.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) for an axis along the axle about which the wheel is rotating. Initially the flywheel has \(30.0 \mathrm{~J}\) of kinetic energy. It is slowing down with an angular acceleration of magnitude \(0.500 \mathrm{rev} / \mathrm{s}^{2} .\) How long does it take for the rotational kinetic energy to become half its initial value, so it is \(15.0 \mathrm{~J} ?\)

A uniform sphere made of modeling clay has radius \(R\) and moment of inertia \(I_{1}\) for rotation about a diameter. It is flattened to a disk with the same radius \(R .\) In terms of \(I_{1},\) what is the moment of inertia of the disk for rotation about an axis that is at the center of the disk and perpendicular to its flat surface?
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