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Chapter 5: Problem 101

A racetrack curve has radius \(90.0 \mathrm{~m}\) and is banked at an angle of \(18.0^{\circ} .\) The coefficient of static friction between the tires and the roadway is \(0.400 .\) A race car with mass \(1200 \mathrm{~kg}\) rounds the curve with the maximum speed to avoid skidding. (a) As the car rounds the curve, what is the normal force exerted on it by the road? What are the car's (b) radial acceleration and (c) speed?

Short Answer

Expert verified
The normal force exerted on the car by the road is 11217N. The car's radial acceleration is 6.22 m/s^2 and its maximum speed is 23.6 m/s.

Step by step solution

01

Calculating Normal Force

First of all, find the components of gravitational force. To do this, break it down into two components: one perpendicular to the road's surface and the other parallel to the surface. The perpendicular component is given by \( mg\cos \theta \), where \( m \) is the mass of the car, \( g \) is the gravitational force and \( \theta \) is the angle of the banked curve. The parallel component of the gravitational force is given by \( mg\sin \theta \). In equilibrium, the total force on the car perpendicular to the road surface should be zero. The perpendicular component of gravitational force is balanced by the normal force. Therefore, the normal force \( F_N \) is \( F_N = mg\cos \theta \). Plugging in the given values, we get \( F_N = 1200kg * 9.8 m/s^2 * \cos18^{\circ} = 11217N \).
02

Calculating Radial Acceleration

The radial acceleration (also called centripetal acceleration) of the car can be determined using the formula: \( a_r = \frac{v^2}{r} \). Here \(v\) is the velocity of the car and \(r\) is the radius of the curve. The maximum velocity before skidding occurs when the friction between the tires and road is maximum. This friction is equal to \( uF_N \), where \( u \) is the coefficient of static friction. The equilibrium in the radial direction implies that the radial acceleration times mass (\(a_r m\)) is equal to the sum of the horizontal (radial) component of gravitational force and friction. Combining these two equations we get: \( u mg \cos \theta = m a_r - mg \sin \theta \) hence \( a_r = g(u \cos \theta + \sin \theta) \). Substituting the given values we get \( a_r = 9.8m/s^2 * (0.4 \cos 18^{\circ} + \sin 18^{\circ}) = 6.22 m/s^2 \).
03

Calculating Maximum Speed

The maximum speed can be found by rearranging the equation for centripetal acceleration: \( v = \sqrt{a_rr} \). After substitution the given values, we get \( v = \sqrt{6.22 m/s^2 * 90.0 m} = 23.6 m/s \).

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Key Concepts

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

Static Friction
Static friction is the force that keeps an object at rest when other forces try to move it. It acts between two surfaces, preventing them from sliding past each other. In this exercise, static friction plays a crucial role in allowing the race car to navigate the banked curve without skidding.

This frictional force depends on two factors:
  • The normal force (perpendicular force) exerted by the road on the tires.
  • The coefficient of static friction, which is a measure of the stickiness between the two surfaces.
The maximum static frictional force can be calculated using the formula: \[ f_s = ext{coefficient of static friction} imes ext{normal force} \]This means it's important to know the normal force, as it directly influences the frictional force that prevents skidding.

When cornering, if the centripetal force, required to keep the car moving in a circle, exceeds this maximum static frictional force, the car will begin to skid. However, as long as the frictional force is greater than or equal to the required centripetal force, the car remains securely on the track.
Centripetal Acceleration
Centripetal acceleration is crucial for understanding how objects move in circular paths. It is the acceleration directed towards the center of the circle, keeping the object moving along the curve. For this exercise, the race car experiences centripetal acceleration as it rounds the track.

The formula for centripetal acceleration is:\[ a_c = \frac{v^2}{r} \]where:
  • \( v \) is the velocity of the car.
  • \( r \) is the radius of the circular path.
Centripetal acceleration is always perpendicular to the velocity of the car because it's directed towards the center of the circle. This is a key requirement for maintaining circular motion.

In the context of this exercise, the balance of forces involves static friction and gravitational forces contributing toward generating enough centripetal acceleration to keep the car on track without skidding. The car鈥檚 speed determines how effectively this balance is achieved.
Normal Force
The normal force is a force exerted by a surface in response to an object resting on it. It acts perpendicular (or "normal") to the surface, which, in this scenario, is the banked curve of the racetrack.

On a flat surface, normal force usually equals the gravitational force exerted by the object (weight). However, on an inclined surface, such as the banked curve described, the normal force changes its magnitude because it's influenced by both gravity and the angle of the incline.

For a banked curve, the normal force is calculated as:\[ F_N = mg\cos(\theta) \]Here,
  • \( m \) is the mass of the race car.
  • \( g \) is the acceleration due to gravity (approximately \(9.8 \text{m/s}^2\)).
  • \( \theta \) is the angle of the bank.
In this exercise, after computing this, the normal force acts as a stabilizing component, helping the car to effectively use static friction to maintain its trajectory and accommodate the necessary centripetal force for staying on the curved path.

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Most popular questions from this chapter

Block \(B,\) with mass \(5.00 \mathrm{~kg},\) rests on block \(A,\) with mass \(8.00 \mathrm{~kg}\), which in turn is on a horizontal tabletop (Fig. P5.92). There is no friction between block \(A\) and the tabletop, but the coefficient of static friction between blocks \(A\) and \(B\) is \(0.750 .\) A light string attached to block \(A\) passes over a friction less, massless pulley, and block \(C\) is suspended from the other end of the string. What is the largest mass that block \(C\) can have so that blocks \(A\) and \(B\) still slide together when the system is released from rest?

\(\mathrm{A} 45.0 \mathrm{~kg}\) crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds \(313 \mathrm{~N}\). Then you must reduce your push to \(208 \mathrm{~N}\) to keep it moving at a steady \(25.0 \mathrm{~cm} / \mathrm{s}\). (a) What are the coefficients of static and kinetic friction between the crate and the floor? (b) What push must you exert to give it an acceleration of \(1.10 \mathrm{~m} / \mathrm{s}^{2} ?\) (c) Suppose you were performing the same experiment on the moon, where the acceleration due to gravity is \(1.62 \mathrm{~m} / \mathrm{s}^{2}\). (i) What magnitude push would cause it to move? (ii) What would its acceleration be if you maintained the push in part (b)?

Friction and Climbing Shoes. Shoes made for the sports of bouldering and rock climbing are designed to provide a great deal of friction between the foot and the surface of the ground. Such shoes on smooth rock might have a coefficient of static friction of 1.2 and a coefficient of kinetic friction of 0.90 . For a person wearing these shoes, what's the maximum angle (with respect to the horizontal) of a smooth rock that can be walked on without slipping? (a) \(42^{\circ} ;\) (b) \(50^{\circ} ;\) (c) \(64^{\circ} ;\) (d) larger than \(90^{\circ}\).

Genesis Crash. On September \(8,2004,\) the Genesis spacecraft crashed in the Utah desert because its parachute did not open. The \(210 \mathrm{~kg}\) capsule hit the ground at \(311 \mathrm{~km} / \mathrm{h}\) and penetrated the soil to a depth of \(81.0 \mathrm{~cm}\). (a) What was its acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) and in \(g\) 's) assumed to be constant, during the crash? (b) What force did the ground exert on the capsule during the crash? Express the force in newtons and as a multiple of the capsule's weight. (c) How long did this force last?

Jack sits in the chair of a Ferris wheel that is rotating at a constant \(0.100 \mathrm{rev} / \mathrm{s}\). As Jack passes through the highest point of his circular path, the upward force that the chair exerts on him is equal to one- fourth of his weight. What is the radius of the circle in which Jack travels? Treat him as a point mass.
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