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What is the angular velocity of a 75.0 -cm-diameter tire on an automobile traveling at \(90.0 \mathrm{km} / \mathrm{h}\) ?

Short Answer

Expert verified
The angular velocity of the 75.0 cm-diameter tire on an automobile traveling at 90.0 km/h is approximately \(16.0 \, \mathrm{rad/s}\).

Step by step solution

01

Convert linear velocity to meters per second

First, we need to convert the linear velocity from km/h to m/s. We will use the conversion factor \(1 \mathrm{km} = 1000 \mathrm{m}\) and \(1 \mathrm{h} = 3600 \mathrm{s}\). The given linear velocity is 90.0 km/h: \(v = 90.0 \frac{\mathrm{km}}{\mathrm{h}} \cdot \frac{1000 \mathrm{m}}{1 \mathrm{km}} \cdot \frac{1 \mathrm{h}}{3600 \mathrm{s}}\)
02

Calculate tire's radius

The diameter of the tire is given as 75.0 cm. To find the radius, we need to divide the diameter by 2 and convert it to meters. We will use the conversion factor \(1 \mathrm{m} = 100 \mathrm{cm}\). \(r = \frac{75.0 \mathrm{cm}}{2} \cdot \frac{1 \mathrm{m}}{100 \mathrm{cm}}\)
03

Relate linear and angular velocities

Now, we can relate the linear and angular velocities using the formula \(v = r \omega\), where \(v\) is the linear velocity, \(r\) is the tire's radius, and \(\omega\) is the angular velocity. \(\omega = \frac{v}{r}\)
04

Calculate the angular velocity

Finally, substitute the values of the linear velocity and radius into the formula for the angular velocity: \(\omega = \frac{90.0 \frac{\mathrm{km}}{\mathrm{h}} \cdot \frac{1000 \mathrm{m}}{1 \mathrm{km}} \cdot \frac{1 \mathrm{h}}{3600 \mathrm{s}}}{\frac{75 \mathrm{cm}}{2} \cdot \frac{1 \mathrm{m}}{100 \mathrm{cm}}}\) Calculate the value of \(\omega\). The angular velocity of the tire is the calculated \(\omega\), in radians per second.

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Key Concepts

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

Linear to Angular Velocity Conversion
Understanding how to convert linear velocity to angular velocity is essential for problems involving rotating objects, such as wheels on a vehicle. Linear velocity refers to the speed of an object traveling in a straight line, while angular velocity denotes how fast the object rotates around a fixed point or axis.

The relationship between linear and angular velocities is expressed through the formula:
\( v = r \omega \), where \( v \) is the linear velocity, \( r \) is the radius of the rotation, and \( \omega \) is the angular velocity. To find the angular velocity, we rearrange the formula to:
\( \omega = \frac{v}{r} \).

This conversion is particularly useful because it allows us to understand the speed at which the angle changes for a point on the edge of a rotating object. For the tire in our exercise, converting its linear motion to angular motion requires this calculation, giving us insight into how quickly the wheel is spinning.
Units of Measurement Conversion
Handling different units of measurement is a fundamental skill in physics and applied sciences. For instance, converting kilometers per hour (km/h) to meters per second (m/s) is vital in problems related to velocity. The conversion process involves using the relationships between units.

For velocity conversion we use:
\( 1 \text{ km} = 1000 \text{ m} \) and \( 1 \text{ h} = 3600 \text{ s} \).

Applying these conversion factors helps standardize units across various physical quantities, making calculations more straightforward and reliable. For circumference-based calculations, such as those involving tire rotations, it's often necessary to convert diameters and radii from centimeters to meters using the factor: \( 1 \text{ m} = 100 \text{ cm} \).

Proper unit conversion ensures that the computed angular velocity is in the universally accepted radians per second (rad/s), making it easy to compare and apply in other equations and scenarios.
Tire Radius Calculation
The radius of a tire is a critical value when determining rotational characteristics like angular velocity. The radius is half the diameter of the tire, an essential step in relating linear distance traveled to the number of rotations.

To calculate the radius from the diameter, we simply divide the diameter by two. If the diameter is given in centimeters (cm), and we require the radius in meters (m) for our calculations, a conversion is also necessary using the factor: \( 1 \text{ m} = 100 \text{ cm} \).

In our exercise, we find the radius of the automobile's tire by halving its diameter of 75.0 cm and then converting this measurement into meters, which are more suitable for calculating angular velocity in the International System of Units (SI).
Velocity and Radius Relationship
The connection between linear velocity and radius plays a fundamental role in understanding rotational movement. The formula \( v = r \omega \) represents this relationship, indicating that linear velocity (\( v \)) is directly proportional to the radius (\( r \)) of a tire and its angular velocity (\( \omega \)).

A larger radius implies that the same angular velocity will result in a higher linear velocity, meaning a point on the circumference travels a greater linear distance in the same amount of time.

In our exercise, by knowing the radius and the speed at which the car travels, we can accurately determine the angular velocity of the tire. Conversely, if we know the angular velocity and the radius, we can determine the speed of the car. This relationship is pivotal in not just physics, but engineering, robotics, and any field involving circular motion.

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Most popular questions from this chapter

The Sun's mass is \(2.0 \times 10^{30} \mathrm{kg}\), its radius is \(7.0 \times 10^{5} \mathrm{km}, \quad\) and \(\quad\) it \(\quad\) has \(\quad\) a rotational period of approximately 28 days. If the Sun should collapse into a white dwarf of radius \(3.5 \times 10^{3} \mathrm{km},\) what would its period be if no mass were ejected and a sphere of uniform density can model the Sun both before and after?

A bug of mass 0.020 kg is at rest on the edge of a solid cylindrical disk \((M=0.10 \mathrm{kg}, R=0.10 \mathrm{m})\) rotating in a horizontal plane around the vertical axis through its center. The disk is rotating at \(10.0 \mathrm{rad} / \mathrm{s}\). The bug crawls to the center of the disk. (a) What is the new angular velocity of the disk? (b) What is the change in the kinetic energy of the system? (c) If the bug crawls back to the outer edge of the disk, what is the angular velocity of the disk then? (d) What is the new kinetic energy of the system? (e) What is the cause of the increase and decrease of kinetic energy?

A cylindrical can of radius \(R\) is rolling across a horizontal surface without slipping. (a) After one complete revolution of the can, what is the distance that its center of mass has moved? (b) Would this distance be greater or smaller if slipping occurred?

A propeller consists of two blades each 3.0 m in length and mass \(120 \mathrm{kg}\) each. The propeller can be approximated by a single rod rotating about its center of mass. The propeller starts from rest and rotates up to \(1200 \mathrm{rpm}\) in 30 seconds at a constant rate. (a) What is the angular momentum of the propeller at \(t=10 \mathrm{s} ; t=20 \mathrm{s} ?\) What is the torque on the propeller?

The blades of a wind turbine are \(30 \mathrm{m}\) in length and rotate at a maximum rotation rate of 20 rev/min. (a) If the blades are \(6000 \mathrm{kg}\) each and the rotor assembly has three blades, calculate the angular momentum of the turbine at this rotation rate. (b) What is the torque require to rotate the blades up to the maximum rotation rate in 5 minutes?

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