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Chapter 6: Problem 54

mechanical engineers must consider how small variations in certain parameters can alter the net force on a passenger. Consider a passenger of mass \(m\) riding around a horizontal circle of radius \(r\) at speed \(v\). What is the variation \(d F\) in the net force magnitude for (a) a variation \(d r\) in the radius with \(v\) held constant, (b) a variation \(d v\) in the speed with \(r\) held constant, and (c) a variation \(d T\) in the period with \(r\) held constant?

Short Answer

Expert verified
(a) \(dF = -\frac{mv^2}{r^2} \cdot dr\), (b) \(dF = \frac{2mv}{r} \cdot dv\), (c) \(dF = -\frac{8\pi^2 mr}{T^3} \cdot dT\).

Step by step solution

01

Expressing Centripetal Force

The centripetal force on a passenger moving around a circle is given by the formula \( F = \frac{mv^2}{r} \). This force keeps the passenger moving in a circle.
02

Variation in Net Force for Radius Change (\(d r\))

Differentiate the expression of force with respect to \(r\) while keeping \(v\) constant. We have:\[ dF = \frac{\partial}{\partial r}\left( \frac{mv^2}{r} \right) \cdot dr = -\frac{mv^2}{r^2} \cdot dr \]
03

Variation in Net Force for Speed Change (\(d v\))

Differentiate the expression of force with respect to \(v\) while keeping \(r\) constant. We have:\[ dF = \frac{\partial}{\partial v}\left( \frac{mv^2}{r} \right) \cdot dv = \frac{2mv}{r} \cdot dv \]
04

Expressing the Period of Circular Motion

The period \(T\) of circular motion is given by \( T = \frac{2\pi r}{v} \). Rearrange to express \(v\) as \( v = \frac{2\pi r}{T} \).
05

Variation in Net Force for Period Change (\(d T\))

Substitute \(v = \frac{2\pi r}{T}\) into the original force equation:\[ F = \frac{m(\frac{2\pi r}{T})^2}{r} = \frac{4\pi^2 mr}{T^2} \]Now, differentiate with respect to \(T\) while keeping \(r\) constant:\[ dF = \frac{\partial}{\partial T}\left( \frac{4\pi^2 mr}{T^2} \right) \cdot dT = -\frac{8\pi^2 mr}{T^3} \cdot dT \]

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Key Concepts

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

Uniform Circular Motion
When an object moves along a circular path at a constant speed, it experiences what we call uniform circular motion. This might sound a bit tricky at first, but think of it like a car going around a curved section of a track at a steady pace.

In this type of motion, the speed or magnitude of the velocity is constant, but the direction is continuously changing. This change in direction is crucial because it means that even though the speed does not change, the velocity does. Remember, velocity includes direction, whereas speed does not.

As a result, there is always an inward pointing force acting on the object that keeps it in that circular path, known as the centripetal force. Imagine tying a ball to a string and spinning it around; the tension in the string is keeping the ball moving in its circle.
  • The magnitude of this force can be calculated using the equation: \( F = \frac{mv^2}{r} \).
  • Here, \(m\) stands for mass, \(v\) is the velocity, and \(r\) is the radius of the circle.
Understanding this simple setup of forces helps in knowing not just how objects move in circles, but also how changing any one of those variables affects the motion.
Differentiation in Physics
Differentiation is a tool used in physics and math to find out how a change in one quantity affects another. In the context of physics, particularly related to calculating variations in centripetal force, differentiation becomes handy.

To differentiate is to find the rate at which a one variable changes with respect to another. Here we'll see how this concept is applied to find variations of force based on the radius and speed. Suppose we have already expressed the centripetal force formula:
  • For a change in radius, keeping speed constant, we find: \[ dF = -\frac{mv^2}{r^2} \cdot dr \]
  • For a change in speed, with constant radius: \[ dF = \frac{2mv}{r} \cdot dv \]
Differentiation offers a powerful way to see how small tweaks in radius and speed can have impacts on the force experienced.

Remember, these differentiated expressions help quantify exact changes in force, letting us know whether the force increases or decreases with these changes.
Newton's Second Law
Newton's Second Law is a foundational concept in physics that ties together force, mass, and acceleration. In its simplest form, the law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (\(F = ma\)).

In the scenario of uniform circular motion, this law is neatly illustrated. Although the object is moving in a circle at constant speed, it is still accelerating because the direction of its velocity is constantly changing. This acceleration towards the center of the circle is the centripetal acceleration.

Newton's second law helps us understand this by setting up that essential equation for centripetal force:
  • The force needed to keep an object moving in a circle (centripetal force) is directly proportional to the square of its speed, inversely proportional to the radius of the circle, and directly proportional to its mass.
The application of Newton's second law in circular motion allows us to solve for unknowns, adjust experiments, and predict the motion of objects given certain conditions.

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Most popular questions from this chapter

An old streetcar rounds a flat corner of radius \(9.1 \mathrm{~m}\), at 16 \(\mathrm{km} / \mathrm{h}\). What angle with the vertical will be made by the loosely hanging hand straps?

A circular curve of highway is designed for traffic moving at \(60 \mathrm{~km} / \mathrm{h}\). Assume the traffic consists of cars without negative lift. (a) If the radius of the curve is \(150 \mathrm{~m}\), what is the correct angle of banking of the road? (b) If the curve were not banked, what would be the minimum coefficient of friction between tires and road that would keep traffic from skidding out of the turn when traveling at \(60 \mathrm{~km} / \mathrm{h} ?\)

ssM A filing cabinet weighing \(556 \mathrm{~N}\) rests on the floor. The coefficient of static friction between it and the floor is \(0.68\), and the coefficient of kinetic friction is \(0.56 .\) In four different attempts to move it, it is pushed with horizontal forces of magnitudes (a) 222 \(\mathrm{N}\), (b) \(334 \mathrm{~N},(\mathrm{c}) 445 \mathrm{~N}\), and \((\mathrm{d}) 556 \mathrm{~N}\). For each attempt, calculate the magnitude of the frictional force on it from the floor. (The cabinet is initially at rest.) (e) In which of the attempts does the cabinet move?

Suppose the coefficient of static friction between the road and the tires on a car is \(0.60\) and the car has no negative lift. What speed will put the car on the verge of sliding as it rounds a level curve of \(30.5 \mathrm{~m}\) radius?

The terminal speed of a sky diver is \(160 \mathrm{~km} / \mathrm{h}\) in the spreadeagle position and \(310 \mathrm{~km} / \mathrm{h}\) in the nosedive position. Assuming that the diver's drag coefficient \(C\) does not change from one position to the other, find the ratio of the effective cross-sectional area \(A\) in the slower position to that in the faster position.
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