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Chapter 6: Problem 58

A car goes around a curve on a road that is banked at an angle of \(33.5^{\circ} .\) Even though the road is slick, the car will stay on the road without any friction between its tires and the road when its speed is \(22.7 \mathrm{m} / \mathrm{s}\). What is the radius of the curve?

Short Answer

Expert verified
The radius of the curve is approximately 79.46 meters.

Step by step solution

01

Understanding the concept of a banked curve without friction

A banked curve allows cars to make turns without relying on friction to provide the necessary centripetal force. When a car moves around such a curve, the component of the normal force provides the centripetal acceleration required for the car to stay in its circular path.
02

Visualize the forces at work

On a banked curve, the car experiences a gravitational force downward, a normal force perpendicular to the surface, and a component of this normal force that acts as centripetal force. Since we're dealing with no friction, this centripetal force is entirely provided by the horizontal component of the normal force.
03

Formulate the equations

For a car traveling at speed \(v\) on a banked road with angle \(\theta = 33.5^{\circ}\), the equation of motion without friction is:\[ \tan(\theta) = \frac{v^2}{rg} \]where \(r\) is the radius of the curve and \(g = 9.81\,\text{m/s}^2\) is the acceleration due to gravity.
04

Solve for the radius \(r\)

Rearranging the equation for \(r\), we get:\[ r = \frac{v^2}{g \tan(\theta)} \]Substitute the given values \(v = 22.7\,\text{m/s}\), \(g = 9.81\,\text{m/s}^2\), and \(\theta = 33.5^{\circ}\):\[ r = \frac{(22.7)^2}{9.81 \times \tan(33.5^{\circ})} \]
05

Calculate the result

Calculate \(\tan(33.5^{\circ})\) using a calculator:\[ \tan(33.5^{\circ}) \approx 0.661 \]Now, substitute this back into the formula:\[ r = \frac{515.29}{9.81 \times 0.661} \approx \frac{515.29}{6.486} \approx 79.46 \, \text{m} \]So the radius of the curve is approximately \(79.46 \, \text{m}\).

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Key Concepts

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

Centripetal Force
Centripetal force is essential for any object moving in a circular path. This force acts towards the center of the circle, keeping the object on its curved trajectory.
In the context of a banked curve, centripetal force is crucial for a car to travel around the curve without slipping off.
The centripetal force needed is provided entirely by the horizontal component of the normal force when there is no friction involved.
Understanding how centripetal force works in these scenarios helps in analyzing and calculating the radius of the curve.
  • Centripetal force is always directed towards the center of the circular path.
  • The magnitude of this force is calculated by the formula: \( F_c = \frac{mv^2}{r} \), where \( m \) is mass, \( v \) is velocity, and \( r \) is the radius of the curve.
Normal Force
Normal force is a contact force that acts perpendicular to the surface of contact. For a car on a banked curve, the normal force is crucial for maintaining balance.
On a banked curve without friction, the normal force splits into components: vertical and horizontal. The horizontal component provides the centripetal force needed for circular motion.
  • The vertical component balances the gravitational force, keeping the car from sliding down.
  • Understanding these components helps decipher the motion of a car on a banked curve in the absence of friction.
Angle of Banking
The angle of banking, denoted usually by \( \theta \), is the angle at which a road or track is inclined compared to the horizontal.
This angle is crucial because it helps vehicles make a turn on a curve by using a component of the normal force to provide the necessary centripetal force.
In the given exercise, the angle is \( 33.5^{\circ} \). This angle ensures that, at the correct speed, a car can make the turn without relying on friction.
  • The steeper the angle, the more effective it is in aiding the turn through the component of the normal force.
  • Proper calculation of this angle is important for safe and efficient transportation in hilly or curvy terrains.
Motion Without Friction
In scenarios involving a banked curve, motion without friction simplifies analysis since the frictional force is absent.
This means we solely rely on gravity and the normal force components to understand the dynamics of objects on the curve.
For this problem, the lack of friction implies that only the horizontal component of the normal force acts as the centripetal force. This is a key consideration in calculating the turn radius.
  • The analysis without friction helps in designing curves that manage forces more predictably for safety in adverse conditions like rain or ice.
  • It also aids in understanding the engineering behind road designs that rely heavily on curve banking rather than material friction.

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Most popular questions from this chapter

As plants grow, they tend to align their stems and roots along the direction of the gravitational field. This tendency, which is related to differential concentrations of plant hormones known as auxins, is referred to as gravitropism. As an illustration of gravitropism, experiments show that seedlings placed in pots on the rim of a rotating turntable do not grow in the vertical direction. Do you expect their stems to tilt inward-toward the axis of rotation -or outward-away from the axis of rotation?

Find the centripetal acceleration at the top of a test tube in a centrifuge, given that the top is \(4.2 \mathrm{cm}\) from the axis of rotation and that its linear speed is \(77 \mathrm{m} / \mathrm{s}\).

A 48-kg crate is placed on an inclined ramp. When the angle the ramp makes with the horizontal is increased to \(26^{\circ}\), the crate begins to slide downward. (a) What is the coefficient of static friction between the crate and the ramp? (b) At what angle does the crate begin to slide if its mass is doubled?

Force Times Time At the local hockey rink, a puck with a mass of \(0.12 \mathrm{kg}\) is given an initial speed of \(v_{0}=6.7 \mathrm{m} / \mathrm{s}\). (a) If the coefficient of kinetic friction between the ice and the puck is 0.13 , how much time \(t\) does it take for the puck to come to rest? (b) If the mass of the puck is doubled, does the frictional force \(F\) exerted on the puck increase, decrease, or stay the same? Explain. (c) Does the stopping time of the puck increase, decrease, or stay the same when its mass is doubled? Explain. (d) For the situation considered in part (a), show that \(F t=m v_{0}\). (The significance of this result will be discussed in Chapter \(9,\) where we will see that \(m v\) is the momentum of an object.)

Pulling up on a rope, you lift a \(4.35-\mathrm{kg}\) bucket of water from a well with an acceleration of \(1.78 \mathrm{m} / \mathrm{s}^{2}\). What is the tension in the rope?
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