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Chapter 5: Problem 81

Your car's wheels are \(65 \mathrm{cm}\) in diameter and the wheels are spinning at an angular velocity of 101 rad/s. How fast is your car moving in kilometers per hour (assume no slippage)?

Short Answer

Expert verified
Question: Given that the diameter of a car's wheels is 65 cm and their angular velocity is 101 rad/s, calculate the speed of the car in kilometers per hour. Answer: The car is moving at a speed of approximately 118.17 km/h.

Step by step solution

01

Write down the provided information

We are given the diameter of the wheels, \(65 \mathrm{cm}\), and the angular velocity of the wheels, \(101 \mathrm{rad/s}\). Our goal is to find the linear velocity of the car in kilometers per hour.
02

Calculate the radius of the wheels

To find the radius of the wheels, divide the diameter by 2: \(radius = \frac{diameter}{2} = \frac{65 \mathrm{cm}}{2} = 32.5 \mathrm{cm}\)
03

Convert the radius to meters

To convert the radius from centimeters to meters, divide by 100: \(radius_m = \frac{32.5 \mathrm{cm}}{100} = 0.325 \mathrm{m}\)
04

Relate angular and linear velocity

The formula relating angular velocity (\(\omega\)) to linear velocity (\(v\)) is: \(v = r\omega\) Plug in the values for the radius and angular velocity: \(v = (0.325 \mathrm{m})(101 \mathrm{rad/s}) = 32.825 \mathrm{m/s}\)
05

Convert meters per second to kilometers per hour

To convert meters per second to kilometers per hour, multiply by \(\frac{3600}{1,000}\): \(v_{km/h} = (32.825 \mathrm{m/s}) \cdot \frac{3600}{1000} = 118.17 \mathrm{km/h}\) So, the car is moving at a speed of approximately \(118.17 \mathrm{km/h}\).

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Most popular questions from this chapter

Objects that are at rest relative to the Earth's surface are in circular motion due to Earth's rotation. (a) What is the radial acceleration of an object at the equator? (b) Is the object's apparent weight greater or less than its weight? Explain. (c) By what percentage does the apparent weight differ from the weight at the equator? (d) Is there any place on Earth where a bathroom scale reading is equal to your true weight? Explain.
A wheel's angular acceleration is constant. Initially its angular velocity is zero. During the first \(1.0-\mathrm{s}\) time interval, it rotates through an angle of \(90.0^{\circ} .\) (a) Through what angle does it rotate during the next 1.0 -s time interval? (b) Through what angle during the third 1.0 -s time interval?
A person rides a Ferris wheel that turns with constant angular velocity. Her weight is \(520.0 \mathrm{N}\). At the top of the ride her apparent weight is \(1.5 \mathrm{N}\) different from her true weight. (a) Is her apparent weight at the top \(521.5 \mathrm{N}\) or \(518.5 \mathrm{N} ?\) Why? (b) What is her apparent weight at the bottom of the ride? (c) If the angular speed of the Ferris wheel is \(0.025 \mathrm{rad} / \mathrm{s},\) what is its radius?
A coin is placed on a turntable that is rotating at 33.3 rpm. If the coefficient of static friction between the coin and the turntable is \(0.1,\) how far from the center of the turntable can the coin be placed without having it slip off?
A biologist is studying plant growth and wants to simulate a gravitational field twice as strong as Earth's. She places the plants on a horizontal rotating table in her laboratory on Earth at a distance of \(12.5 \mathrm{cm}\) from the axis of rotation. What angular speed will give the plants an effective gravitational field \(\overrightarrow{\mathrm{g}}_{\mathrm{eff}},\) whose magnitude is \(2.0 \mathrm{g} ?\) \([\) Hint: Remember to account for Earth's gravitational field as well as the artificial gravity when finding the apparent weight. \(]\)
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