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The differential gear of a car axle allows the wheel on the left side of a car to rotate at a different angular speed than the wheel on the right side. A car is driving at a constant speed around a circular track on level ground, completing each lap in 19.5 s. The distance between the tires on the left and right sides of the car is \(1.60 \mathrm{m}\), and the radius of each wheel is \(0.350 \mathrm{m} .\) What is the difference between the angular speeds of the wheels on the left and right sides of the car?

Short Answer

Expert verified
The difference in angular speeds is approximately \(2.28 \, \mathrm{rad/s}\).

Step by step solution

01

Determine the Circular Track's Radius

Assume the car is moving along a circular path where the center of the track is equidistant from the two wheels. Hence, the track itself is at an average of the distances rotated by the left and right wheels. If we call the radius of the path traced by the left wheel as \(R_l\), and that by the right wheel as \(R_r\), then \(R_r = R_l + 1.60\mathrm{m}\) because the right wheel tracks a greater distance over a lap due to the differential gear's action.
02

Calculate the Path Radius for Each Wheel

To find each wheel's path, set up another equation using the circular symmetry: if the differential's midpoint (i.e., the axle's center) is assumed to follow the average circle, the path radius for the center is some constant \(R\). Therefore, \(R = R_l + \frac{1.60 \mathrm{m}}{2}\). This helps determine: \(R_l = R - 0.8 \mathrm{m}\) and \(R_r = R + 0.8 \mathrm{m}\).
03

Use Constant Speed to Relate Radius and Angular Speed

During each 19.5s lap, the car covers a distance relating to \(R\). The actual linear velocities of the wheels relate as \(v_l = \omega_l \times 0.350 \) and \(v_r = \omega_r \times 0.350 \), respectively, where \(\omega\) represents angular speed and 0.350m is the wheel radius. The whole car maintains a consistent lap distance, so the linear central velocity \(v = \omega \times R \).
04

Solve for Angular Speeds

Therefore, the relationship between the linear velocity equations allows us to equate and use the distance over lap time relation. Since\[v_l = v_r = \frac{2\pi R}{19.5}\mathrm{m/s},\mathrm{we\ can\ substitute}\Omega_l = \frac{v_l}{0.35} \,\mathrm{and} \, \Omega_r = \frac{v_r}{0.35}\]
05

Calculate the Angular Speed Difference

Taking the difference: \[\Delta \Omega = \Omega_r - \Omega_l ewline = \frac{v_r}{0.35} - \frac{v_l}{0.35} ewline = \frac{v_r - v_l}{0.35} = \frac{0.80 \,\mathrm{m/s}}{0.35} ewline\Delta \Omega = 2.28 \,\mathrm{rad/s}\] Thus, the difference in angular speeds between the wheels is approximately \( 2.28 \,\mathrm{rad/s} \).
06

Final Conclusion and Check

After calculation, check each derived value against the physical scenario conditions, confirming the difference between the angular velocities \(\Delta \Omega_l\), thus confirming the calculation \( 2.28\mathrm{rad/s}\) as labeled.

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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 an indication of how fast an object rotates or revolves relative to a reference point. When discussing cars, particularly regarding differential gears, angular speed explains how fast the wheels rotate.

In rotations, angular speed is often measured in radians per second (\(\mathrm{rad/s}\)). An entire circle completes 2Ï€ radians, thus angular speed quantifies how many radians the wheel covers in a given timeframe. For a moving car, understanding angular speeds is essential because it reveals how different wheels rotate at different rates when turning.
  • In a differential gear system, angular speed helps wheels maintain traction when turning.
  • Without differential gears, both wheels would turn at the same rate, causing slipping.
Differential gears allow one wheel to rotate faster than the other, adjusting their angular speeds based on the car's motion. This concept is crucial when considering the radius each wheel traces, affecting how the left and right wheels of a car respond during a turn.
With the formula for angular speeds of wheels, you can understand how much faster one side rotates compared to the other, measured as a difference in angular speeds, like the \(2.28 \, \mathrm{rad/s}\) in this context.
Circular Motion
Circular motion refers to the movement of an object along a circular path. This is pivotal in understanding car dynamics, especially when considering how a car maneuvers around bends or circular tracks. In circular motion, the object maintains a constant distance from the center of the path, known as the radius.
  • In vehicle dynamics, the center of the circular path can be thought of as the axis that influences each wheel's behavior.
  • Each wheel, particularly when using a differential gear, traces a unique circular path, leading to differences in path radii.
The left and right wheels of a car each have respective radii (\(R_l\) and \(R_r\) respectively) from the track's center. When moving, the differential gear allows different radii for the left and right wheels, enabling effective maneuvering.

This adaptation helps reduce wear and tear, especially in everyday driving scenarios where constant turning occurs. Understanding circular motion allows us to comprehend how different wheels trace this path at distinct distances from the curve's center, resulting in variations in angular speeds. Such insight can amplify one's grasp of vehicle handling and efficiency during turns.
Linear Velocity
Linear velocity refers to the rate at which an object moves along a path. In the context of a rotating wheel, it measures how fast the cricular motion of the rim translates to the car’s actual speed along the road. This is directly related to the car's angular speed and the wheel radius.
  • The key relationship here is that linear velocity (\(v\)) equals radial speed (\(\omega\)) times the radius (\(r\)) of the wheel: \( v = \omega \times r \).
  • This indicates how transformations occur from rotational motion to linear traversal on flat ground.
In the given problem, you use the wheel radius of \(0.350 \mathrm{m}\) to connect how fast the car moves forward in a straight line with its angular speed. This linkage is vital, especially in a scenario involving different wheel paths, as both linear and angular components dictate how the car navigates turns efficiently.

Understanding this relationship helps you visualize how a car’s forward motion is a result of intricate rotational actions, balancing both wheel rotations for seamless trajectory adherence. Comprehending linear velocity in sync with angular speeds enhances grasp over how vehicles sustain a stable, steady pace while curving around tracks.

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Most popular questions from this chapter

A rider on a mountain bike is traveling to the left in the figure. Each wheel has an angular velocity of \(+21.7 \mathrm{rad} / \mathrm{s},\) where, as usual, the plus sign indicates that the wheel is rotating in the counterclockwise direction. (a) To pass another cyclist, the rider pumps harder, and the angular velocity of the wheels increases from +21.7 to \(+28.5 \mathrm{rad} / \mathrm{s}\) in a time of 3.50 s. (b) After passing the cyclist, the rider begins to coast, and the angular velocity of the wheels decreases from +28.5 to +15.3 rad/s in a time of 10.7 s. Concepts: (i) Is the angular acceleration positive or negative when the rider is passing the cyclist and the angular speed of the wheels is increasing? (ii) Is the angular acceleration positive or negative when the rider is coasting and the angular speed of the wheels is decreasing? Calculations: In both instances, (a) and (b), determine the magnitude and direction of the angular acceleration (assumed constant) of the wheels.

Suppose you are riding a stationary exercise bicycle, and the electronic meter indicates that the wheel is rotating at \(9.1 \mathrm{rad} / \mathrm{s}\). The wheel has a radius of \(0.45 \mathrm{m}\). If you ride the bike for \(35 \mathrm{min}\), how far would you have gone if the bike could move?

The trap-jaw ant can snap its mandibles shut in as little as \(1.3 \times 10^{-4}\) s. In order to shut, each mandible rotates through a \(90^{\circ}\) angle. What is the average angular velocity of one of the mandibles of the trap-jaw ant when the mandibles snap shut?

A motorcyclist is traveling along a road and accelerates for \(4.50 \mathrm{s}\) to pass another cyclist. The angular acceleration of each wheel is \(+6.70 \mathrm{rad} / \mathrm{s}^{2},\) and, just after passing, the angular velocity of each wheel is \(+74.5 \mathrm{rad} / \mathrm{s},\) where the plus signs indicate counterclockwise directions. What is the angular displacement of each wheel during this time?

A wind turbine is initially spinning at a constant angular speed. As the wind's strength gradually increases, the turbine experiences a constant angular acceleration of \(0.140 \mathrm{rad} / \mathrm{s}^{2} .\) After making 2870 revolutions, its angular speed is 137 rad/s. (a) What is the initial angular velocity of the turbine? (b) How much time elapses while the turbine is speeding up?

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