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Chapter 39: Problem 1031

An elliptical cam rotates about its focus \(\mathrm{F}\), causing a roller at P move up and down parallel to the \(\mathrm{y}\) -axis. If the diameters of the cam are six and ten inches and it rotates at the rate of 240 r.p.m., how fast is the roller moving at the moment when the long axis of the cam makes an angle of \(60^{\circ}\) with the line of motion of the roller?

Short Answer

Expert verified
At the given moment when the long axis of the cam makes an angle of \(60^{\circ}\) with the line of motion of the roller, the roller is moving downwards at a rate of \(-12\pi\) inches per second.

Step by step solution

01

Establish the equation of the ellipse

An ellipse with its center at the origin is defined by the equation: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) In this case, the semi-major axis a is half of the length of the long diameter (10 inches) and the semi-minor axis b is half of the length of the short diameter (6 inches). Therefore, a = 5 and b = 3. The equation of the ellipse is: \(\frac{x^2}{25} + \frac{y^2}{9} = 1\)
02

Find the position of the roller in terms of x and y

As the cam rotates about focus F, the position of roller P will lie on the ellipse. At the given moment, the angle 胃 between the long axis of the cam and the line of motion of the roller is 60 degrees. If 伪 is the angle between the x-axis and the line connecting the origin and P, we have: \(\alpha = 60^{\circ} + 胃\) Now, we can express the position of P in terms of 伪: \(x = a \cos{\alpha}\) \(y = b \sin{\alpha}\)
03

Differentiate the equations with respect to time

To find the roller's velocity in the y direction, we need to differentiate the equation for y with respect to time t: \(\frac{dy}{dt} = b \cos{\alpha} \cdot \frac{d\alpha}{dt}\) Also, we know that the cam rotates at 240 r.p.m, or 4 revolutions per second. To convert this into d伪/dt, we multiply by 2蟺 (one complete revolution) and 4 (revolutions per second): \(\frac{d\alpha}{dt} = 4 \cdot 2\pi = 8\pi \,\text{rad/s}\)
04

Calculate the velocity of the roller

Now we can plug in the values for b, 伪, and d伪/dt into the equation for dy/dt: \(\frac{dy}{dt} = 3 \cos{(60^{\circ} + 60^{\circ})} \cdot 8\pi\) \(\frac{dy}{dt} = 3 \cos{120^{\circ}} \cdot 8\pi\) \(\frac{dy}{dt} = -\frac{3}{2} \cdot 8\pi\) \(\frac{dy}{dt} = -12\pi \,\text{in/s}\) At this moment, the roller is moving downwards (negative sign) at a rate of \(12\pi\) inches per second.

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Key Concepts

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

Elliptical Motion
Elliptical motion refers to the path of an object moving in the shape of an ellipse, which is an oval contour. In classical mechanics, elliptical motion is usually associated with celestial bodies orbiting under gravity, but it can also apply to mechanical systems like cams in engines or machinery. An ellipse has two focal points, and the sum of the distances from any point on the ellipse to these foci is constant.

In our problem, we consider an elliptical cam that rotates about one of its foci. Due to the elliptical shape, the motion of a roller following the cam varies in speed as the cam turns. At different points on the ellipse, the distance from the focus changes, and so does the vertical motion speed of the roller connected to the cam. Understanding elliptical motion is crucial because it allows us to apply concepts from geometry and calculus to find related rates of change, such as the speed of the roller at any given moment.
Related Rates
In calculus, related rates are used to determine how two or more quantities that change over time are connected to one another. To find a related rate, one typically uses the chain rule to relate the rates of change of the quantities involved. We often deal with situations where one quantity's rate of change depends on another's. For example, if one quantity changes faster, it might cause the other to change faster as well.

In the problem at hand, we're asked to determine the velocity of a roller as an elliptical cam rotates, which is a classic related rates problem. Since the roller's vertical speed is related to the angular speed of the cam and the angle of rotation, we use differentiation to find the relationship between these variables. The key step is to set up an equation that connects all the related quantities, differentiating with respect to time to expose how the rate of one quantity affects the rate of another. This concept reveals that geometric constraints, like those imposed by the shape of an ellipse, can create complex relationships between rates that calculus can unravel.
Differentiation of Trigonometric Functions
Differentiating trigonometric functions is a fundamental skill in calculus, which involves finding the rate of change, or derivative, of sine, cosine, tangent, and other trigonometric functions with respect to an angle. These derivatives are essential in solving problems involving periodic motion or wave phenomena, where angles and their trigonometric functions naturally occur.

In our example, after setting up the position of the roller in terms of the sine and cosine of the angle 伪, we differentiate these positions with respect to time, applying the chain rule to the trigonometric functions. The differentiation reveals how the change in the roller's position is affected by both the change in the cam's rotation angle and time. This step is crucial in related rates problems involving trigonometric functions, such as the movement of objects following a circular or elliptical path.

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Most popular questions from this chapter

A particle moves along the curve: \(\mathrm{y}^{2}=\mathrm{x}^{3}\). If its horizontal velocity is constant and equal to \(2 \mathrm{ft} / \mathrm{sec} .\), find (a) the angular velocity (w) and angular acceleration (a) and (b) \(\mathrm{w}\) and \(\alpha\) at the point \((1,1)\).

If the path of a particle is a curve with an inflection point, show that the normal component of acceleration vanishes at such a point. Illustrate with the curve: \(\mathrm{x}=\mathrm{t}, \mathrm{y}=\mathrm{t}^{3}\).

A particle starts at the origin and travels along the curve: \(\mathrm{y}=\mathrm{x}^{2}\). As it passes through the point \((3,9)\), its velocity is such that the \(\mathrm{x}\) -component, \(\mathrm{V}_{\mathrm{x}}=\mathrm{D}_{\mathrm{t}} \mathrm{x}\), is \(2 \mathrm{ft} / \mathrm{sec}\). Given that \(D_{x} y=\left[\left(D_{t} y\right) /\left(D_{t} x\right)\right]\), find the \(y\) -component: \(V_{y}=D_{t} y\), and the resultant speed \(\left.\mathrm{V}=\sqrt{(} \mathrm{V}^{2} \mathrm{x}+\mathrm{V}^{2}{ }_{y}\right)\). Use the \(\Delta-\) method.

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A helicopter flying horizontally in a straight line at 60 miles an hour and an elevation of 1760 feet crosses a straight, level road at right angles. At the same time, a car passes underneath at 30 miles per hour. One minute later, how far apart are the two, and at what rate do they separate?
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