/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 When a toy car is rapidly scoote... [FREE SOLUTION] | 91Ó°ÊÓ

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When a toy car is rapidly scooted across the floor, it stores energy in a flywheel. The car has mass 0.180 \(\mathrm{kg}\) , and its flywheel has moment of inertia \(4.00 \times 10^{-5} \mathrm{kg} \cdot \mathrm{m}^{2} .\) The car is 15.0 \(\mathrm{cm}\) long. An advertisement claims that the car can travel at a scale speed of up to 700 \(\mathrm{km} / \mathrm{h}(440 \mathrm{mi} \mathrm{h}) .\) The scale speed is the speed of the toy car multiplied by the ratio of the length of an actual car to the length of the toy. Assume a length of 3.0 \(\mathrm{m}\) for a real car. (a) For a scale speed of 700 \(\mathrm{km} / \mathrm{h}\) , what is the actual translational speed of the car? (b) If all the kinetic energy that is initially in the flywheel is converted to the translational kinetic energy of the toy, how much energy is originally stored in the flywheel? (c) What initial angular velocity of the flywheel was needed to store the amount of energy calculated in part (b)?

Short Answer

Expert verified
The actual speed of the toy car is 35 km/h. The energy stored in the flywheel is approximately 3.15 J. The initial angular velocity of the flywheel is approximately 397.5 rad/s.

Step by step solution

01

Calculate the Speed Ratio

The scale speed is based on a ratio between the length of an actual car and the toy car. Calculate this ratio as follows: \[ \text{Speed Ratio} = \frac{\text{Real Car Length}}{\text{Toy Car Length}} = \frac{3.0 \text{ m}}{0.15 \text{ m}} = 20 \]
02

Calculate Actual Translational Speed

The actual speed of the toy car can be found by dividing the scale speed by the speed ratio. Given the scale speed is 700 km/h, the actual speed is calculated as:\[ \text{Actual Speed} = \frac{700 \text{ km/h}}{20} = 35 \text{ km/h} \]
03

Determine Translational Kinetic Energy

Using the formula for translational kinetic energy, \( KE_{\text{trans}} = \frac{1}{2}m v^2 \), where \( m = 0.180 \text{ kg} \) and \( v = 35 \text{ km/h} = \frac{35}{3.6} \text{ m/s} \): \[ KE_{\text{trans}} = \frac{1}{2} \times 0.180 \times \left(\frac{35}{3.6}\right)^2 \approx 3.15 \text{ J} \]
04

Convert Energy to Rotational Energy

Assuming all the energy is converted from the flywheel energy, the initial energy stored in the flywheel \( KE_{\text{rot}} = KE_{\text{trans}} \): \[ KE_{\text{rot}} = 3.15 \text{ J} \]
05

Calculate Initial Angular Velocity

Use the formula for rotational kinetic energy, \( KE_{\text{rot}} = \frac{1}{2}I\omega^2 \), solving for \( \omega \) gives:\[ 3.15 = \frac{1}{2} \times 4.00 \times 10^{-5} \times \omega^2 \]Rearranging gives \( \omega^2 = \frac{2 \times 3.15}{4.00 \times 10^{-5}} \), \[ \omega = \sqrt{\frac{6.3}{4.00 \times 10^{-5}}} \approx 397.5 \text{ rad/s} \]

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Key Concepts

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

Translational kinetic energy
Understanding translational kinetic energy starts with visualizing motion in a straight line. When objects move across a surface, like a toy car scooting across the floor, they possess this type of energy.
The formula to find translational kinetic energy is \( KE_{\text{trans}} = \frac{1}{2}mv^2 \), where:
  • \( m \) is the mass of the object.
  • \( v \) is the velocity of the object.
Translational kinetic energy depends directly on both mass and speed. In our problem, the car's mass is \( 0.180 \text{ kg} \) and its speed is converted to meters per second to apply the formula effectively. When you plug these numbers in, you calculate how much energy the toy car stores simply by moving across the floor. This principle helps us understand how energy is transferred in motion and is essential in topics ranging from basic physics to complex engineering.
Rotational energy
Rotational energy, also known as the energy of an object due to its rotation, plays a significant role in our flywheel example.
Flywheels are designed to store energy through spinning, based on the object's moment of inertia and its angular velocity.
The formula to calculate rotational energy is \( KE_{\text{rot}} = \frac{1}{2}I\omega^2 \), where:
  • \( I \) is the moment of inertia, which measures how much resistance an object has to changes in its rotational state.
  • \( \omega \) is the angular velocity, or how fast the object spins around an axis.
In the problem, all the energy stored in the rotating flywheel converts into translational kinetic energy of the toy car as it moves. This conversion illustrates the conservation of energy principle, indicating that energy can change forms but not be created or destroyed. Understanding this relationship helps urban planners, engineers, and physicists alike in developing more efficient systems.
Angular velocity calculation
Finding angular velocity involves understanding how fast an object spins around its axis.
In our problem, we know the energy stored in the flywheel and the moment of inertia, which allows us to find the angular velocity. The formula to calculate angular velocity when rotational kinetic energy is known is derived from \( KE_{\text{rot}} = \frac{1}{2}I\omega^2 \).
Rewriting this equation, we solve for \( \omega \):
  • \( \omega^2 = \frac{2 \times KE_{\text{rot}}}{I} \)
  • \( \omega = \sqrt{\frac{2 \times 3.15 \text{ J}}{4.00 \times 10^{-5} \text{ kg} \cdot \text{m}^2}} \)
Calculation produces an angular velocity of around 397.5 rad/s. It reflects how rapidly the flywheel spins to store the needed amount of energy that eventually translates into the car's motion. This concept is fundamental in devices like turbines and engines, where rotational motion is converted to power various applications.

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Most popular questions from this chapter

A wheel of diameter 40.0 \(\mathrm{cm}\) starts from rest and rotates with a constant angular acceleration of 3.00 \(\mathrm{rad} / \mathrm{s}^{2}\) . At the instant the wheel has computed its second revolution, compute the radial acceleration of a point on the rim in two ways: (a) using the relationship \(a_{\mathrm{rad}}=\omega^{2} r\) and \((\mathrm{b})\) from the relationship \(a_{\mathrm{red}}=v^{2} / r\)

A straight piece of reflecting tape extends from the center of a wheel to its rim. You darken the room and use a camera and strobe unit that flashes once every 0.050 s to take pictures of the wheel as it rotates counterclockwise. You trigger the strobe so that the first flash \((t=0)\) occurs when the tape is horizontal to the right at an angular displacement of zero. For the following situations draw a sketch of the photo you will get for the time exposure over five flashes (at \(t=0,0.050 \mathrm{s}, 0.100 \mathrm{s}, 0.150 \mathrm{s},\) and 0.200 \(\mathrm{s} )\) and graph \(\theta\) versus \(t\) and \(\omega\) versus \(t\) for \(t=0\) to \(t=0.200 \mathrm{s}\) (a) The angular velocity is constant at 10.0 rev \(/ \mathrm{s}\) . (b) The wheel starts from rest with a constant angular acceleration of 25.0 rev \(/ \mathrm{s}^{2}\) . (c) The wheel is rotating at 10.0 rev \(/ \mathrm{s}\) at \(t=0\) and changes angular velocity at a constant rate of \(-50.0 \mathrm{rev} / \mathrm{s}^{2}\) .

The spin cycles of a washing machine have two angular speeds, 423 rev/min and 640 rev/min. The internal diameter of the drum is 0.470 \(\mathrm{m}\) . (a) What is the ratio of the maximum radial force on the laundry for the higher angular speed to that for the lower speed? (b) What is the ratio of the maximum tangential speed of the laundry for the higher angular speed to that for the lower speed? (c) Find the laundry's maximum tangential speed and the maxi- mum radial acceleration, in terms of \(g .\)

Trip to Mars. You are working on a project with NASA to launch a rocket to Mars, with the rocket blasting off from earth when earth and Mars are aligned along a straight line from the sun. If Mars is now \(60^{\circ}\) shead of earth in its orbit around the sun, when should you launch the rocket? (Note: All the planets orbit the sun in the same direction, 1 year on Mars is 1.9 earth- years, and assume circular orbits for both planets.)

Compact Disc. A compact disc (CD) stores music in a coded pattern of tiny pits \(10^{-7} \mathrm{m}\) deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 \(\mathrm{mm}\) and 58.0 \(\mathrm{mm}\) , respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 \(\mathrm{m} / \mathrm{s}\) . (a) What is the angular speed of the CD when the innermost part of the track is scanned? The outermost part of the track? (b) The maximum playing time of a \(\mathrm{CD}\) is 74.0 min. What would be the length of the track on such a maximum-duration \(\mathrm{CD}\) if it were stretched out in a straight line? (c) What is the average angular acceleration of a maximum-duration CD during its 74.0 -min playing time? Take the direction of rotation of the dise to be positive.

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