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The Nature of Science and Physics

Kinematics

Introduction to One-Dimensional KinematicsDisplacementVectors, Scalars, and Coordinate SystemsTime, Velocity, and SpeedVideo: One-Dimensional KinematicsAccelerationMotion Equations for Constant Acceleration in One DimensionProblem-Solving Basics for One-Dimensional KinematicsFalling ObjectsGraphical Analysis of One-Dimensional Motion

Two-Dimensional Kinematics

Dynamics: Force and Newton's Laws of Motion

Introduction to Dynamics: Newton's Laws of MotionDevelopment of Force ConceptNewton's First Law of Motion: InertiaNewton's Second Law of Motion: Concept of a SystemNewton's Third Law of Motion: Symmetry in ForcesVideo: Newton's LawsNormal, Tension, and Other Examples of ForcesProblem-Solving StrategiesFurther Applications of Newton's Laws of MotionExtended Topic: The Four Basic Forces—An Introduction

Further Applications of Newton's Laws: Friction, Drag, and Elasticity

Uniform Circular Motion and Gravitation

Introduction to Uniform Circular Motion and GravitationRotation Angle and Angular VelocityCentripetal AccelerationCentripetal ForceFictitious Forces and Non-inertial Frames: The Coriolis ForceNewton's Universal Law of GravitationVideo: GravitationSatellites and Kepler's Laws: An Argument for Simplicity

Work, Energy, and Energy Resources

Introduction to Work, Energy, and Energy ResourcesWork: The Scientific DefinitionKinetic Energy and the Work-Energy TheoremGravitational Potential EnergyVideo: Potential and Kinetic EnergyConservative Forces and Potential EnergyNonconservative ForcesConservation of EnergyPowerWork, Energy, and Power in HumansWorld Energy Use

Linear Momentum and Collisions

Rotational Motion and Angular Momentum

Introduction to Rotational Motion and Angular MomentumAngular AccelerationKinematics of Rotational MotionVideo: Rotational MotionDynamics of Rotational Motion: Rotational InertiaRotational Kinetic Energy: Work and Energy RevisitedAngular Momentum and Its ConservationVideo: Angular MomentumCollisions of Extended Bodies in Two DimensionsGyroscopic Effects: Vector Aspects of Angular Momentum

Statics and Torque

Fluid Statics

Introduction to Fluid StaticsWhat Is a Fluid?DensityPressureVariation of Pressure with Depth in a FluidPascal's PrincipleGauge Pressure, Absolute Pressure, and Pressure MeasurementArchimedes' PrincipleVideo: BuoyancyCohesion and Adhesion in Liquids: Surface Tension and Capillary ActionPressures in the Body

Fluid Dynamics and Its Biological and Medical Applications

Introduction to Fluid Dynamics and Biological and Medical ApplicationsFlow Rate and Its Relation to VelocityBernoulli's EquationVideo: Fluid FlowThe Most General Applications of Bernoulli's EquationViscosity and Laminar Flow; Poiseuille's LawThe Onset of TurbulenceMotion of an Object in a Viscous FluidMolecular Transport Phenomena: Diffusion, Osmosis, and Related Processes

Temperature, Kinetic Theory, and the Gas Laws

Heat and Heat Transfer Methods

Thermodynamics

Introduction to ThermodynamicsThe First Law of ThermodynamicsThe First Law of Thermodynamics and Some Simple ProcessesIntroduction to the Second Law of Thermodynamics: Heat Engines and Their EfficiencyCarnot's Perfect Heat Engine: The Second Law of Thermodynamics RestatedApplications of Thermodynamics: Heat Pumps and RefrigeratorsEntropy and the Second Law of Thermodynamics: Disorder and the Unavailability of EnergyStatistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation

Oscillatory Motion and Waves

Introduction to Oscillatory Motion and WavesHooke's Law: Stress and Strain RevisitedPeriod and Frequency in OscillationsSimple Harmonic Motion: A Special Periodic MotionVideo: Harmonic MotionThe Simple PendulumEnergy and the Simple Harmonic OscillatorUniform Circular Motion and Simple Harmonic MotionDamped Harmonic MotionForced Oscillations and ResonanceWavesSuperposition and InterferenceEnergy in Waves: Intensity

Physics of Hearing

- Calculate coefficient of friction on a car tire.
- Calculate ideal speed and angle of a car on a turn.

Any net force causing uniform circular motion is called a

By using the expressions for centripetal acceleration

$a_c=\frac{v^2}{r};a_c=r\omega^2\\$

, we get two expressions for the centripetal force F$\text{F}_c=m\frac{v^2}{r};\text{F}_c=mr\omega^2\\$

.You may use whichever expression for centripetal force is more convenient. Centripetal force

Note that if you solve the first expression for

$\displaystyle{r}=\frac{mv^2}{\text{F}_c}\\$

.This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.

- Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s.
- Assuming an unbanked curve, find the minimum static coefficient of friction, between the tires and the road, static friction being the reason that keeps the car from slipping (see Figure 2).

$\text{F}_c=\frac{mv^2}{r}\\$

. Thus,$\displaystyle\text{F}_c=\frac{mv^2}{r}=\frac{\left(900\text{ kg}\right)\left(25.0\text{ m/s}\right)^2}{\left(500\text{ m}\right)}=1125\text{ N}\\$

.
F_{c}* = f* = *μ*_{s}*N = μ_{s}mg.*

$\begin{cases}\text{F}_c=m\frac{v^2}{r}\\\text{F}_c=mr\omega^2\end{cases},\text{ }m\frac{v^2}{r}=\mu_s{mg}\\$

We solve this for $\displaystyle\mu_s=\frac{v^2}{rg}\\$

.
$\displaystyle\mu_s=\frac{\left(25.0\text{ m/s}\right)^2}{\left(500\text{ m}\right)\left(9.80\text{ m/s}^2\right)}=0.13\\$

.
(Because coefficients of friction are approximate, the answer is given to only two digits.)$\begin{cases}\text{F}_c=m\frac{v^2}{r}\\\text{F}_c=mr\omega^2\end{cases}\\$

because Let us now consider

For

Figure 3 shows a free body diagram for a car on a frictionless banked curve. If the angle

$N\sin\theta=\frac{mv^2}{r}\\$

.
Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction. From the figure, we see that the vertical component of the normal force is Now we can combine the last two equations to eliminate

$N=\frac{mg}{\cos\theta}\\$

, and substituting this into the first yields$\displaystyle\begin{matrix}\\mg\frac{\sin\theta}{\cos\theta}=\frac{mv^2}{r}\\mg\tan\left(\theta\right)=\frac{mv^2}{r}\\\tan\theta=\frac{v^2}{rg}\end{matrix}\\$

Taking the inverse tangent gives$\theta=\tan^{-1}\left(\frac{v^2}{rg}\right)\\$

(ideally banked curve, no friction).
This expression can be understood by considering how $\tan\theta=\frac{v^2}{rg}\\$

, we get $\begin{matrix}\\v=\left[\left(100 m\right)\left(9.80\text{ m/s}^2\right)\left(2.14\right)\right]^{1/2}\\\text{ }=45.8\end{matrix}\\$

Calculations similar to those in the preceding examples can be performed for a host of interesting situations in which centripetal force is involved—a number of these are presented in this chapter’s Problems and Exercises.

- Centripetal force F
is any force causing uniform circular motion. It is a "center-seeking" force that always points toward the center of rotation. It is perpendicular to linear velocity_{c}*v*and has magnitude F, which can also be expressed as_{c}= ma_{c}

$\begin{cases}\text{F}_c=m\frac{v^2}{r}\\\text{or}\\\text{F}_c=mr\omega^2\end{cases}\\$

- If you wish to reduce the stress (which is related to centripetal force) on high-speed tires, would you use large- or small-diameter tires? Explain.
- Define centripetal force. Can any type of force (for example, tension, gravitational force, friction, and so on) be a centripetal force? Can any combination of forces be a centripetal force?
- If centripetal force is directed toward the center, why do you feel that you are ‘thrown’ away from the center as a car goes around a curve? Explain.
- Race car drivers routinely cut corners as shown in Figure 7. Explain how this allows the curve to be taken at the greatest speed.

- A number of amusement parks have rides that make vertical loops like the one shown in Figure 8. For safety, the cars are attached to the rails in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will supply the centripetal force. What other force acts and what is its direction if: (a) The car goes over the top at faster than this speed? (b) The car goes over the top at slower than this speed?

- What is the direction of the force exerted by the car on the passenger as the car goes over the top of the amusement ride pictured in Figure 8 under the following circumstances: (a) The car goes over the top at such a speed that the gravitational force is the only force acting? (b) The car goes over the top faster than this speed? (c) The car goes over the top slower than this speed?
- As a skater forms a circle, what force is responsible for making her turn? Use a free body diagram in your answer.
- Suppose a child is riding on a merry-go-round at a distance about halfway between its center and edge. She has a lunch box resting on wax paper, so that there is very little friction between it and the merry-go-round. Which path shown in Figure 9 will the lunch box take when she lets go? The lunch box leaves a trail in the dust on the merry-go-round. Is that trail straight, curved to the left, or curved to the right? Explain your answer.

- Do you feel yourself thrown to either side when you negotiate a curve that is ideally banked for your car’s speed? What is the direction of the force exerted on you by the car seat?
- Suppose a mass is moving in a circular path on a frictionless table as shown in figure. In the Earth’s frame of reference, there is no centrifugal force pulling the mass away from the centre of rotation, yet there is a very real force stretching the string attaching the mass to the nail. Using concepts related to centripetal force and Newton’s third law, explain what force stretches the string, identifying its physical origin.

- (a) A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force must she exert to stay on if she is 1.25 m from its center? (b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at 3.00 rev/min if she is 8.00 m from its center? (c) Compare each force with her weight.
- Calculate the centripetal force on the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev/s. Assume the mass is 4 kg.
- What is the ideal banking angle for a gentle turn of 1.20 km radius on a highway with a 105 km/h speed limit (about 65 mi/h), assuming everyone travels at the limit?
- What is the ideal speed to take a 100 m radius curve banked at a 20.0° angle?
- (a) What is the radius of a bobsled turn banked at 75.0° and taken at 30.0 m/s, assuming it is ideally banked? (b) Calculate the centripetal acceleration. (c) Does this acceleration seem large to you?
- Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure 4. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components—friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system’s weight). (a) Show that
*θ*(as defined in the figure) is related to the speed*v*and radius of curvature*r*of the turn in the same way as for an ideally banked roadway—that is,$\theta=\tan^{-1}\frac{v^2}{rg}\\$; (b) Calculate*θ*for a 12.0 m/s turn of radius 30.0 m (as in a race).

- A large centrifuge, like the one shown in Figure 5a, is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries. (a) At what angular velocity is the centripetal acceleration 10 g if the rider is 15.0 m from the center of rotation? (b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in Figure 5b. At what angle
*θ*below the horizontal will the cage hang when the centripetal acceleration is 10 g? (Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free body diagram of the forces to see what the angle*θ*should be.)

**Integrated Concepts.**If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads). (a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0º. (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?- Modern roller coasters have vertical loops like the one shown in Figure 6. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 15.0 m and the downward acceleration of the car is 1.50 g?

**Unreasonable Results.**(a) Calculate the minimum coefficient of friction needed for a car to negotiate an unbanked 50.0 m radius curve at 30.0 m/s. (b) What is unreasonable about the result? (c) Which premises are unreasonable or inconsistent?

3. 4.14º

5. (a) 24.6 m; (b) 36.6 m/s

7. (a) 2.56 rad/s; (b) 5.71º

8. (a) 16.2 m/s; (b) 0.234

10. (a) 1.84; (b) A coefficient of friction this much greater than 1 is unreasonable; (c) The assumed speed is too great for the tight curve.