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Chapter 7: Problem 140

A circular road of radius \(1,000 \mathrm{~m}\) has a banking angle of \(45^{\circ}\). What will be the maximum safe speed (in \(\mathrm{m} / \mathrm{s}\) ) of a car whose mass is \(2,000 \mathrm{~kg}\) and the coefficient of friction between the tyre and the road is \(0.5\) a. 172 b. 124 c. 99 d. 86

Short Answer

Expert verified
The maximum safe speed is 99 m/s.

Step by step solution

01

Understand the Forces Involved

In a banked curve, several forces act on the car: gravitational force pulling it down, normal force from the road, and frictional force that can act in the up or down direction along the slope. We must consider the car's centripetal force needed to keep it moving in a circle.
02

Define the Equations of the Forces

The forces can be expressed using formulas: - Centripetal force needed is \( F_c = \frac{mv^2}{r} \), where \( m \) is mass, \( v \) is velocity, and \( r \) is radius.- Gravitational force is \( F_g = mg \).- Normal force \( F_n \) and frictional force \( F_f = \mu F_n \) act perpendicular to each other on the banked surface.
03

Set Up the Equilibrium Equations

Using the bank angle \( \theta = 45^{\circ} \), the normal and frictional forces provide part of the centripetal force:- Horizontal components: \( F_n \sin\theta + \mu F_n \cos\theta = \frac{mv^2}{r} \).- Vertical components: \( F_n \cos\theta - \mu F_n \sin\theta = mg \). Where \( \mu \) is the friction coefficient, \( g \) is gravitational acceleration (\(9.8 \mathrm{~m/s}^2\)).
04

Solve for Normal Force

From the vertical equilibrium: \[ F_n ( \cos\theta - \mu \sin\theta) = mg \]Solve for \( F_n \): \[ F_n = \frac{mg}{\cos\theta - \mu \sin\theta} \]
05

Substitute and Solve for Maximum Speed

Substitute \( F_n \) into the horizontal equilibrium:\[ F_n \sin\theta + \mu F_n \cos\theta = \frac{mv^2}{r} \]Substitute \( F_n = \frac{mg}{\cos\theta - \mu \sin\theta} \) into the centripetal equation:\[ \left(\frac{mg}{\cos\theta - \mu \sin\theta}\right)\sin\theta + \mu \left(\frac{mg}{\cos\theta - \mu \sin\theta}\right)\cos\theta = \frac{mv^2}{r} \]Simplify and solve for \( v \) to find the maximum speed.
06

Evaluating and Simplification

Calculate: - \( \sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}} \)- Substitute into the equation:\[ v = \sqrt{r g \frac{\sin\theta + \mu \cos\theta}{\cos\theta - \mu \sin\theta}} \]\[ v = \sqrt{1000 \times 9.8 \times \frac{\frac{1}{\sqrt{2}} + 0.5 \times \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} - 0.5 \times \frac{1}{\sqrt{2}}}} \]After simplification, \( v \approx 99 \text{ m/s} \).

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Key Concepts

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

Centripetal Force: The Key to Circular Motion
When a car moves around a circular road, it relies on centripetal force to keep it on its path. But what exactly is centripetal force? Let鈥檚 break it down. In simple terms, centripetal force is a force that acts on an object moving in a circular path, directed towards the center of the circle. This force is essential for maintaining the circular motion and is calculated using the formula \[ F_c = \frac{mv^2}{r} \] where \( m \) is the mass of the object, \( v \) is the velocity, and \( r \) is the radius of the circle.
  • The greater the mass or velocity, the greater the centripetal force needed.
  • This force is not a separate "real" force; rather, it is the result of other forces acting on the body, such as gravity or friction.
In our scenario, the car's weight and the friction between its tires and the road provide this necessary force so the car doesn't skid outward.
Banking Angle: Balancing Forces for Stability
Imagine driving around a perfectly flat curve. The car would have to rely solely on friction to avoid slipping. Not ideal in all situations. This is where a banking angle becomes a game-changer.
The banking angle is how much the road tilts towards the curve's center. It's like leaning inward when you run around a corner. The tilt helps direct part of the gravitational force to work with the centripetal force, reducing dependence on friction. The formula used is:
  • Forces are split into horizontal (centripetal) and vertical components.
  • The angle helps balance these forces to help maintain the car's path without slipping.
The formula to find the balance is impacted by the angle 胃, as in the car's case of \(45^{\circ}\). A greater banking angle can allow higher speeds without extra friction, proving crucial for safety, especially on high-speed roads.
Frictional Force: The Invisible Grip
Frictional force is the unsung hero on the road, providing the grip necessary for control. It acts opposite the direction of potential slippage and is calculated as:\[ F_f = \mu F_n \] where \( \mu \) is the coefficient of friction, and \( F_n \) is the normal force.
  • High friction means stronger grip, helping prevent skids on curves.
  • It is influenced by surface conditions and tire quality.
In the exercise, friction works alongside the banking angle to uphold the car's motion in a circle. It's especially vital when speed or road conditions increase the chance of slipping. The coefficient of friction, given as 0.5, plays a pivotal role in determining the maximum safe speed, making sure the tires stick to the road under different forces.

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Most popular questions from this chapter

A wooden block of mass \(M\) resting on a rough horizontal floor is pulled with a force \(F\) at an angle \(\phi\) with the horizontal. If \(\mu\) is the coefficient of kinetic friction between the block and the surface, then acceleration of the block is a. \(\frac{F}{M}(\cos \phi-\mu \sin \phi)-\mu g\) b. \(\frac{\mu F}{M} \cos \phi\) c. \(\frac{F}{M}(\cos \phi+\mu \sin \phi)-\mu g\) d. \(\frac{F}{M} \sin \phi\)

A person is drawing himself up and a trolley on which he stands with some acceleration. Mass of the person is more than the mass of the trolley. As the person increases his force on the string, the normal reaction between person and the trolley will a. increase b. decrease c. remain same d. cannot be predicted as data is insufficient

A wooden block of mass \(M\) resting on a rough horizontal floor is pulled with a force \(F\) at an angle \(\phi\) with the horizontal. If \(\mu\) is the coefficient of kinetic friction between the block and the surface, then acceleration of the block is a. \(\frac{F}{M} \sin \phi\) b. \(\frac{F}{M}(\cos \phi+\mu \sin \phi)-\mu g\) c. \(\frac{\mu F}{M} \cos \phi\) d. \(\frac{F}{M}(\cos \phi-\mu \sin \phi)-\mu g\)

Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A \(15 \mathrm{~kg}\) weight is attached to the rope at the mid point which how no. longer remains horizontal. The minimum tension required to completely straighten the rope is a. \(15 \mathrm{~kg}\) b. \(15 / 2 \mathrm{~kg}\) c. \(5 \mathrm{~kg}\) d. Infinitely large

A triangular prism of mass \(M\) with a block of mass \(m\) placed on it is released from rest on a smooth inclined olane of inclination \(\theta\). The block does not slip on the prism. Then a. The acceleration of the prism is \(g \cos \theta\). b. The acceleration of the prism is \(g \tan \theta\). c. The minimum coefficient of friction between the block and the prism is \(\mu_{\operatorname{mia}}=\cot \theta\). d. The minimum coefficient of friction between the block and the prism is \(\mu_{\min }=\tan \theta\).
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