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An automobile has a total mass of \(1700 \mathrm{~kg}\). It accelerates from rest to \(40 \mathrm{~km} / \mathrm{h}\) in \(10 \mathrm{~s}\). Assume each wheel is a uniform \(32 \mathrm{~kg}\) disk. Find, for the end of the \(10 \mathrm{~s}\) interval, (a) the rotational kinetic energy of each wheel about its axle, (b) the total kinetic energy of each wheel, and (c) the total kinetic energy of the automobile.

Short Answer

Expert verified
(a) 988.36 J, (b) 2966.07 J, (c) 108787.29 J

Step by step solution

01

Convert Velocity to Meters per Second

The automobile accelerates to a final speed of \(40 \, \text{km/h}\). We need to convert this speed to meters per second for further calculations: \[ 40 \, \text{km/h} = \frac{40 \, \text{km}}{1 \, \text{h}} \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} = 11.11 \, \text{m/s} \]
02

Calculate the Linear Velocity of the Wheel's Edge

The automobile's wheels rotate as it moves. The linear speed of the car is equal to the linear speed at the edge of each wheel. If \( v \) is the linear speed of the edge, then:\[ v = 11.11 \, \text{m/s} \]
03

Calculate the Rotational Inertia of the Wheel

For a uniform disk, the rotational inertia \( I \) is given by \( I = \frac{1}{2} m r^2 \), where \( m = 32 \, \text{kg} \) is the mass of the wheel. However, we need radius \( r \) to proceed, which is not provided, so assume: \( r = 0.3 \, \text{m} \) for calculation purposes.
04

Calculate the Angular Velocity of the Wheel

The angular velocity \( \omega \) can be calculated using \( v = r \omega \) where \( v = 11.11 \, \text{m/s} \). Hence,\[ \omega = \frac{v}{r} = \frac{11.11}{0.3} = 37.03 \, \text{rad/s} \]
05

Determine the Rotational Kinetic Energy of Each Wheel

The rotational kinetic energy \( KE_{\text{rot}} \) of each wheel is given by \( KE_{\text{rot}} = \frac{1}{2} I \omega^2 \). With \( I = \frac{1}{2} \times 32 \times (0.3)^2 \mathrm{~kg \cdot m^2} = 1.44 \, \text{kg m}^2 \) and \( \omega = 37.03 \, \text{rad/s} \), we have:\[ KE_{\text{rot}} = \frac{1}{2} \times 1.44 \times (37.03)^2 = 988.36 \, \text{J} \]
06

Calculate the Total Kinetic Energy of Each Wheel

Each wheel has translational kinetic energy in addition to rotational kinetic energy. The translational kinetic energy \( KE_{\text{trans}} \) is given by:\[ KE_{\text{trans}} = \frac{1}{2} m v^2 = \frac{1}{2} \times 32 \times (11.11)^2 = 1977.71 \, \text{J} \]Thus, total kinetic energy of each wheel is:\[ KE_{\text{total ext{,} wheel}} = KE_{\text{rot}} + KE_{\text{trans}} = 988.36 + 1977.71 = 2966.07 \, \text{J} \]
07

Calculate the Total Kinetic Energy of the Automobile

The kinetic energy of the entire automobile includes the translational kinetic energy of the main body plus the kinetic energy of the wheels. The main body weighs \( 1700 - 4 \times 32 = 1572 \, \text{kg} \).Translational kinetic energy for the body is:\[ KE_{\text{body}} = \frac{1}{2} \times 1572 \times (11.11)^2 = 96923.01 \, \text{J} \]Total kinetic energy for wheels is:\[ 4 \times 2966.07 = 11864.28 \, \text{J} \]Overall total kinetic energy of the automobile:\[ KE_{\text{total}} = KE_{\text{body}} + 11864.28 = 108787.29 \, \text{J} \]

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Key Concepts

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

Rotational Kinetic Energy
Rotational kinetic energy is a type of kinetic energy associated with the rotation of an object. It's part of something spinning around an axis. Imagine a wheel or a fan blade, which spins around its center. The energy it possesses due to its rotational motion is called rotational kinetic energy.

For a solid object like a wheel which we consider in this problem, rotational kinetic energy is calculated using the formula: \[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \]where:
  • \( I \) is the rotational inertia (or moment of inertia) of the object, which determines how easy or difficult it is for the object to change its rotational motion.
  • \( \omega \) is the angular velocity in radians per second, telling us how fast the object is spinning.
Understanding this energy helps when calculating the total energy of systems involving rotating parts, like wheels on a car, as in the exercise.
Translational Kinetic Energy
Translational kinetic energy refers to the kinetic energy of an object as it moves in a straight line. Think about a car driving down a road. The energy it uses to propel itself along that path in a straight direction is its translational kinetic energy.

The formula to calculate translational kinetic energy is:\[ KE_{\text{trans}} = \frac{1}{2} mv^2 \]where:
  • \( m \) is the mass of the object in kilograms.
  • \( v \) is the velocity of the object in meters per second.
In the context of a car, every part including the body and wheels contribute to the overall translational kinetic energy. For each wheel in the exercise, it has its own translational energy due to their linear movement in addition to any rotation.
Angular Velocity
Angular velocity is a measure of how quickly something spins around a circular path or its axis. When you think about how fast a wheel turns, you're considering its angular velocity.

Angular velocity, denoted by \( \omega \), is measured in radians per second. It quantifies the rotational speed by capturing the angle through which something rotates in a certain time. The relation between the linear speed \( v \) at the edge of a wheel and angular velocity is given by: \[ v = r \omega \]where:
  • \( v \) is the linear velocity of the wheel's edge.
  • \( r \) is the radius of the wheel.
In the exercise, angular velocity helps in connecting the rotation of the wheels to the linear motion of the car.
Rotational Inertia
Rotational inertia, also known as the moment of inertia, is crucial when considering how objects spin. It's like the rotational equivalent of mass for linear motion.

Rotational inertia \( I \) determines how difficult it is to change an object's rotation. It depends not just on the object's mass, but also on how that mass is distributed relative to the axis of rotation. For a wheel or disk, the rotational inertia can be calculated by:\[ I = \frac{1}{2} m r^2 \]where:
  • \( m \) is the mass of the object.
  • \( r \) is the radius of the wheel.
The exercise involves a disk model of the wheel, which is a common assumption for simplifying calculations. Rotational inertia plays a key role in calculating rotational kinetic energy, hence influencing the total energetic behavior of rotating systems.

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Most popular questions from this chapter

A uniform thin rod of length \(0.500 \mathrm{~m}\) and mass \(4.00 \mathrm{~kg}\) can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a \(3.00 \mathrm{~g}\) bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle \(\theta=60.0^{\circ}\) with the rod (Fig. 11-50). If the bullet lodges in the rod and the angular velocity of the rod is \(10 \mathrm{rad} / \mathrm{s}\) immediately after the collision, what is the bullet's speed just before impact?

A \(140 \mathrm{~kg}\) hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of \(0.150 \mathrm{~m} / \mathrm{s}\). How much work must be done on the hoop to stop it?

Force \(\vec{F}=(-8.0 \mathrm{~N}) \hat{\mathrm{i}}+(6.0 \mathrm{~N}) \hat{\mathrm{j}}\) acts on a particle with position vector \(\vec{r}=(3.0 \mathrm{~m}) \hat{\mathrm{i}}+(4.0 \mathrm{~m}) \hat{\mathrm{j}}\). What are (a) the torque on the particle about the origin, in unit-vector notation, and (b) the angle between the directions of \(\vec{r}\) and \(\vec{F}\) ?

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A uniform solid ball rolls smoothly along a floor, then up a ramp inclined at \(15.0^{\circ} .\) It momentarily stops when it has rolled \(1.50\) \(\mathrm{m}\) along the ramp. What was its initial speed?

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