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On a banked race track, the smallest circular path on which cars can move has a radius of \(112 \mathrm{m},\) while the largest has a radius of \(165 \mathrm{m},\) as the drawing illustrates. The height of the outer wall is \(18 \mathrm{m}\). Find \((\mathrm{a})\) the smallest and (b) the largest speed at which cars can move on this track without relying on friction.

Short Answer

Expert verified
The smallest speed is approximately 20.3 m/s and the largest speed is approximately 25.5 m/s.

Step by step solution

01

Identify the Problem

We need to find the speeds at which cars can move on the banked track without relying on friction. We have radius values for two paths and the height of the outer wall.
02

Clarify Concepts and Variables

In this problem, the banking angle is a vital component. We need to find the speeds relying on the centripetal force stemming from the gravitational component along the banked surface. We denote radius of the track as \(r\), gravitational acceleration as \(g=9.81\ \text{m/s}^2\), and determine angles and speeds separately for both given radii.
03

Derive Formula for Banking Angle

Firstly, using geometry, find the banking angle. The tangent of the angle \(\theta\) is given by \(\tan \theta = \frac{\text{height of wall}}{\text{difference in radii}} = \frac{18}{165 - 112}\). Then calculate \(\theta\).
04

Calculate Banking Angle

Substitute into the formula: \(\tan \theta = \frac{18}{53}\) to compute \(\theta = \tan^{-1} \left( \frac{18}{53} \right)\). Compute the angle with these values.
05

Find the Smallest Speed (Radius = 112 m)

For frictionless banking, the speed formula is: \(v = \sqrt{r \cdot g \cdot \tan \theta}\). Substitute \(r = 112\ \text{m}, g = 9.81\ \text{m/s}^2, \theta\) determined into the equation.
06

Calculate Smallest Speed

Solve \(v = \sqrt{112 \cdot 9.81 \cdot \tan(\theta)}\), substituting the \(\theta\) calculated to find \(v\).
07

Find the Largest Speed (Radius = 165 m)

Use the same speed equation: \(v = \sqrt{r \cdot g \cdot \tan \theta}\), but this time for the radius 165 m. Substitute \(r = 165\ \text{m}\).
08

Calculate Largest Speed

Solve \(v = \sqrt{165 \cdot 9.81 \cdot \tan(\theta)}\), using the previously calculated angle \(\theta\) to find \(v\).

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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 the force that keeps an object moving in a circular path. It acts toward the center of the circle. When a car moves on a banked track, centripetal force is crucial because it ensures that the car stays on its circular path without skidding outwards or inwards. The formula for centripetal force is given by:
  • \( F_c = \frac{mv^2}{r} \)
where \( F_c \) is the centripetal force, \( m \) is the mass of the object, \( v \) is the velocity, and \( r \) is the radius of the circular path.
On a banked track without friction, gravity and the normal force provide the centripetal force needed. The component of gravitational force along the track substitutes the role of friction, which we'll see in an upcoming section.
Banking Angle
The banking angle, often denoted as \( \theta \), is the angle at which the track is inclined relative to the horizontal plane. In our exercise, the banking angle helps generate the needed centripetal force through gravitational components.
This angle is determined by the geometry of the track. To find the banking angle, use the tangent function:
  • \( \tan \theta = \frac{\text{height of the wall}}{\text{difference in radii}} \)
For the given problem, using the measurements, we calculated the banking angle \( \theta = \tan^{-1} \left( \frac{18}{165 - 112} \right) \).
This angle ensures that even without friction, the car can move safely on its path at the calculated speeds.
Frictionless Motion
Frictionless motion in the context of a banked track implies that the car relies solely on the inclination of the track and gravity to maintain its path. Here, the normal force and the gravitational force's vertical component align to create the centripetal force.
For frictionless travel, the speed must match the theoretical speed calculated using the formula:
  • \( v = \sqrt{r \cdot g \cdot \tan \theta} \)
where \( v \) is the speed, \( r \) is the radius, \( g \) is the gravitational acceleration, and \( \tan \theta \) is from the banking angle.
Achieving this velocity means that the car can safely navigate the track without relying on any additional frictional support.
Radius
In circular motion, the radius is the distance from the center of the circle to any point on its circumference. It plays a significant role in determining the needed centripetal force and, subsequently, the velocity at which a car can safely travel on a banked track.
For our scenario, there are two radii: 112 m (smallest path) and 165 m (largest path). These radii represent different paths a car can take on the track.
The value of the radius directly impacts the velocity calculation. A larger radius results in a higher speed permissible by the car, as seen in the calculations for both paths using the centripetal force formula provided.
Gravitational Acceleration
Gravitational acceleration is a constant force acting on objects due to Earth's gravity. Denoted by \( g \), its standard value is \( 9.81 \ \text{m/s}^2 \). This value impacts how forces work on the banked track, providing the component of centripetal force needed in frictionless conditions.
Gravity not only pulls objects downward but also plays a key role in allowing the car to maintain its circular path on the banked track. When calculating speed for frictionless motion, \( g \) is used alongside the radius and banking angle in the formula:
  • \( v = \sqrt{r \cdot g \cdot \tan \theta} \)
This demonstrates how gravitational acceleration contributes to the speed at which the car can safely travel.

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Most popular questions from this chapter

Computer-controlled display screens provide drivers in the Indianapolis 500 with a variety of information about how their cars are performing. For instance, as a car is going through a turn, a speed of \(221 \mathrm{mi} / \mathrm{h}(98.8 \mathrm{m} / \mathrm{s})\) and centripetal acceleration of \(3.00 \mathrm{g}\) (three times the acceleration due to gravity) are displayed. Determine the radius of the turn (in meters).

A satellite circles the earth in an orbit whose radius is twice the earth's radius. The earth's mass is \(5.98 \times 10^{24} \mathrm{kg}\), and its radius is \(6.38 \times 10^{6} \mathrm{m}\). What is the period of the satellite?

The large blade of a helicopter is rotating in a horizontal circle. The length of the blade is \(6.7 \mathrm{m},\) measured from its tip to the center of the circle. Find the ratio of the centripetal acceleration at the end of the blade to that which exists at a point located \(3.0 \mathrm{m}\) from the center of the circle.

Two newly discovered planets follow circular orbits around a star in a distant part of the galaxy. The orbital speeds of the planets are determined to be \(43.3 \mathrm{km} / \mathrm{s}\) and \(58.6 \mathrm{km} / \mathrm{s} .\) The slower planet's orbital period is 7.60 years. (a) What is the mass of the star? (b) What is the orbital period of the faster planet, in years?

There is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted so that it can be rotated about its axis by turning a hand crank. The outer wall of the cylinder is perforated with small holes. You put the wet leaves in the container and turn the crank to spin off the water. The radius of the container is \(12 \mathrm{cm}\) When the cylinder is rotating at 2.0 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?

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