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Chapter 3: Problem 139

The road way bridge over a canal is in the form of an arc of a circle of radius \(20 \mathrm{~m}\). What is the minimum speed with which a car can cross the bridge without leaving contact with the ground at the highest point \(\left(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)\) (a) \(7 \mathrm{~m} / \mathrm{s}\) (b) \(14 \mathrm{~m} / \mathrm{s}\) (c) \(289 \mathrm{~m} / \mathrm{s}\) (d) \(5 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The minimum speed a car can cross the bridge without losing contact with the ground at the highest point is \(14 m/s\). So, the correct answer is (b) \(14 m/s\).

Step by step solution

01

Identify the Essential Variables

The radius of the circular bridge is given as \(20m\), and the acceleration due to gravity (denoted as \(g\)) is known to be \(9.8 m/s^2\). The goal is to find the minimum speed required for a moving car (which we'll label as \(v\)) to remain in contact with the bridge at its highest point.
02

Apply the Principle of Circular Motion

In the case of circular motion, the necessary centripetal force for the car to stay in contact with the bridge is provided by gravitational force. At the highest point these two forces are equal. The equation describing this is \(mg = mv^2/r\), where \(m\) is mass of the car, \(g\) is acceleration due to gravity, \(r\) is radius of the circle, and \(v\) is speed of the car.
03

Simplify the Equation

Both sides of the equation have \(m\), the mass, which cancels out. This simplifies the equation to \(g = v^2/r\).
04

Solve for unknown variable

Our goal is to get \(v\), the velocity, by itself. The equation can be further rearranged to solve for \(v\), i.e., \(v = \sqrt{gr}\). Given \(g = 9.8 m/s^2\) and \(r = 20 m\), substituting these values into the equation we get \(v = \sqrt{9.8*20}\).
05

Compute the Numerical Answer

Using a calculator or computation software, solve for \(v\) by taking the square root of \(9.8*20\), which equals \(14 m/s\).

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Key Concepts

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

Centripetal Force
When an object moves in a circle, it constantly changes direction, meaning it undergoes acceleration. This type of acceleration requires a net force, called centripetal force, which always points towards the center of the circle. In simpler terms, centripetal force is what keeps the object moving in its circular path.
Centripetal force is calculated using the formula:
  • \( F_c = \frac{mv^2}{r} \)
where:
  • \( F_c \) is the centripetal force
  • \( m \) is the mass of the object
  • \( v \) is the velocity of the object
  • \( r \) is the radius of the circle
Understanding centripetal force is crucial because it helps explain why vehicles, like a car crossing a circular bridge, need enough speed to maintain contact with the ground. Without adequate force to keep it on its circular path, the car could lose contact, especially at points where gravity plays a dominant role.
Gravitational Force
The gravitational force is the force of attraction between two objects with mass. On Earth, this force acts downwards towards the planet's center. It is calculated using the equation:
  • \( F_g = mg \)
where:
  • \( F_g \) is the gravitational force
  • \( m \) is the mass of the object
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \) on Earth
In circular motion, such as a car on a bridge, gravity provides the necessary centripetal force to keep the car in contact with the bridge. At the highest point of the circular path, gravity must be strong enough to equal the centripetal force required for the car to maintain contact. By setting these two forces equal to each other, we can determine the minimum velocity needed for the situation.
Velocity Calculation
Velocity is the speed of an object in a specific direction. In our scenario, we aim to calculate the minimum velocity of a car at the highest point of a circular bridge so that it stays in contact with the road.
To find this, we use the simplified equation from balancing centripetal and gravitational forces:
  • \( v = \sqrt{gr} \)
where:
  • \( g \) is the gravitational acceleration (\( 9.8 \text{ m/s}^2 \))
  • \( r \) is the radius of the circle (given as \( 20 \text{ m} \) in this problem)
Plugging in these values, we compute:
  • \( v = \sqrt{9.8 \times 20} \)
  • Solving this gives \( v \approx 14 \text{ m/s} \)
This calculation shows that the minimum speed required for the car to stay on the bridge, without losing contact at the highest point, is \( 14 \text{ m/s} \). This ensures that the forces are in balance and the car remains safely on its path.

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Most popular questions from this chapter

A car sometimes overturns while taking a turn. When it overturns, it is (a) The inner wheel which leaves the ground first (b) The outer wheel which leaves the ground first (c) Both the wheels leave the ground simultaneously (d) Either wheel leaves the ground first

A bucket tied at the end of a \(1.6 \mathrm{~m}\) long string is whirled in a vercar cmere min constant sped. What should be the minimum speed so that the water from the bucket does not spill, when the bucket is at the highest position (Take \(g=10 \mathrm{~m} / \mathrm{sec}^{2}\) ) (a) \(4 \mathrm{~m} / \mathrm{sec}\) (b) \(6.25 \mathrm{~m} / \mathrm{sec}\) (c) \(16 \mathrm{~m} / \mathrm{sec}\) (d) None of these

A stone of mass \(m\) is tied to a string and is moved in a vertical circle of radius \(r\) making \(n\) revolutions per minute. The total tension in the string when the stone is at its lowest point is (a) \(m\left\\{g+\left(\pi^{2} n^{2} r\right) / 900\right\\}\) (b) \(m\left(g+\pi n r^{2}\right)\) (c) \(m(g+\pi n r)\) (d) \(m\left(g+n^{2} r^{2}\right)\)

A mass is supported on a frictionless horizontal surface. It is attached to a string and rotates about a fixed centre at an angular velocity \(\omega_{b}\). If the length of the string and angular velocity are doubled, the tension in the string which was initially \(T_{0}\) is now [AIIMS \(\left.198_{5}\right]\) (a) \(T_{0}\) (b) \(T_{o} / 2\) (c) \(4 T_{0}\) (d) \(8 T_{0}\)

The radius of curvature of a road at a certain turn is \(50 \mathrm{~m}\). The width of the road is \(10 \mathrm{~m}\) and its outer edge is \(1.5 \mathrm{~m}\) higher than the inner edge. The safe speed for such an inclination will be (a) \(6.5 \mathrm{~m} / \mathrm{s}\) (b) \(8.6 \mathrm{~m} / \mathrm{s}\) (c) \(8 \mathrm{~m} / \mathrm{s}\) (d) \(10 \mathrm{~m} / \mathrm{s}\)
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