/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 89 If the banking angle of curved r... [FREE SOLUTION] | 91Ó°ÊÓ

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If the banking angle of curved road is given by \(\tan 1\left(\frac{3}{5}\right)\) and the radius of curvature of the road is \(6 \mathrm{~m}\), then the safe driving speed should not exceed: \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (a) \(86.4 \mathrm{~km} / \mathrm{h}\) (b) \(43.2 \mathrm{~km} / \mathrm{h}\) (c) \(21.6 \mathrm{~km} / \mathrm{h}\) (d) \(30.4 \mathrm{~km} / \mathrm{h}\)

Short Answer

Expert verified
The safe driving speed should not exceed 21.6 km/h.

Step by step solution

01

Understanding the Problem

We are given a curved road with a banking angle and need to find the maximum safe speed to drive along the curve. The banking angle is given by \( \tan^{-1}\left(\frac{3}{5}\right) \), the radius \( R = 6 \) meters, and gravitational acceleration \( g = 10 \) m/s\(^2\).
02

Determine the Banking Angle

The angle \( \theta \) of the banked road is calculated using the inverse tangent function: \( \theta = \tan^{-1}\left(\frac{3}{5}\right) \). This implies that the tangent of the angle is \( \tan\theta = \frac{3}{5} \).
03

Formula for Safe Driving Speed

The formula to find the safe driving speed \( v \) on a banked road, without relying on friction, is \( v = \sqrt{gR\tan\theta} \).
04

Substitute Known Values

Substitute the known values into the formula: \( g = 10 \) m/s\(^2\), \( R = 6 \) m, and \( \tan\theta = \frac{3}{5} \). Therefore, the equation becomes: \[ v = \sqrt{10 \times 6 \times \frac{3}{5}} \].
05

Calculate the Safe Speed

Perform the calculation: \( v = \sqrt{10 \times 6 \times \frac{3}{5}} = \sqrt{36} = 6 \) m/s. Convert this speed from m/s to km/h by multiplying by \( 3.6 \), which gives \( 6 \times 3.6 = 21.6 \) km/h.

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Key Concepts

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

Banking Angle
Banking angle refers to the inclination of a road or track surface, especially during curves, to assist vehicles in making safe and efficient turns. When a road is banked, the outer edge of the curve is elevated compared to the inner edge. This angle is crucial for maintaining control and stability of vehicles, as it helps counteract the centrifugal force that pushes a vehicle outward during a turn. For our problem, the banking angle is determined using the inverse tangent function, specifically, given by \( \theta = \tan^{-1}\left(\frac{3}{5}\right) \). This means the angle itself is calculated based on the ratio between certain forces, represented here as fractions of 3 and 5. Understanding this angle is key to calculating the safe driving speed, as the greater the banking angle, the easier it is for vehicles to navigate the curve safely at higher speeds without relying heavily on friction.
Radius of Curvature
The radius of curvature is a measure that describes the extent to which a road bends. In simpler terms, it is the radius of an imaginary circle that fits exactly into the curve of the road. The significance of the radius in determining safe driving speed is paramount. A smaller radius implies a sharper curve, which typically requires a lower speed to navigate safely. Conversely, a larger radius means a gentler curve, allowing for higher speeds. In the given exercise, the radius of curvature is 6 meters, indicating a moderate curve. This informs the calculation of the maximum safe speed the vehicle can maintain on the curve. The radius directly influences the centripetal force needed to keep the vehicle on the desired path, which ties into the formula for finding the safe speed.
Gravitational Acceleration
Gravitational acceleration, denoted as \( g \), is the acceleration due to gravity experienced by all objects near the Earth's surface. Its standard value is approximately 9.81 m/s², but in the given exercise, it is approximated to 10 m/s² for simplicity. Gravitational acceleration plays a crucial role in different equations of motion and dynamics especially when discussing the impact on vehicles moving along curved paths. For banked roads, it contributes to the inward force needed to keep a vehicle safely on the curve. This inward force is part of what helps counter the natural tendency of the car to slide outward during a turn. Applying its value, 10 m/s² in our specific formula, aids in calculating the forces in play and determining the safe speed that a vehicle can maintain without risking slipping off the intended path of travel.
Tangential Velocity on Curved Paths
Tangential velocity describes the speed of an object moving along a path tangent to a curve. In the context of driving on banked roads, it refers to the speed at which a vehicle drives along the edge of the curve. When studying curves, particularly banked ones, this speed needs to be neither too slow nor too fast to ensure safety and efficiency. The formula for calculating the maximum safe tangential speed \( v \) on a banked road is \( v = \sqrt{gR\tan\theta} \). This equation shows the direct connection between tangential velocity, gravitational force, the curvature of the road, and the banking angle. For the given exercise, substituting the known values of \( g = 10 \) m/s², \( R = 6 \) m, and \( \tan\theta = \frac{3}{5} \) leads to the calculation of a safe velocity of 6 m/s. Converting this to a more practical unit for driving, we get 21.6 km/h, ensuring drivers have a clear understanding of speed limits on these kinds of roads.

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Most popular questions from this chapter

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\(\Lambda\) chain of mass \(m\) and length \(l\) is held vertical, such that its lower end just touches the floor. I released from rest. Find the force exeried by the chain on the table when upper end is about to hit the foor. Solution Force \(F\) exerted by chain consists of two components (a) \(F_{1}\) weight of the fallen portion of the chain, (b) \(F_{2}\) thrust of the falling part of chain. Now consider an clement of chain of length \(d y\) at a height \(y\) from the floor. It will strike the floor with a velocity \(v-\sqrt{2 g y}\). Thus we have, \(\Gamma_{1}=\lambda y g\) Here \(\lambda\) is the mass per unit length of chain and \(\Gamma_{2}-v_{\mathrm{rel}} \frac{d m}{d t}\) We have \(v_{\mathrm{rel}}=v \quad\) and \(\quad d m=\lambda d x \quad \therefore F=-v \frac{\lambda d x}{d t}-\lambda v^{2}\) 'Ihe force exerted by chain on the floor,$$ F=F_{1}+F_{2}=\lambda y g+\lambda v^{2}-\lambda y g+\lambda(\sqrt{2 g y})^{2}=\lambda y g+2 \lambda g=3 \lambda y g $$ When upper end is about to hit the floor, \(y=l\) \(\therefore \quad F=3 \lambda / g=3 m g\)

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