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Chapter 6: Problem 35

Drinking a milkshake The truncated conical container shown here is full of strawberry milkshake that weighs 4\(/ 9\) oz \(/ \mathrm{in} .^{3}\) As you can see, the container is 7 in. deep, 2.5 in. across at the base, and 3.5 in. across at the top (a standard size at Brigham's in Boston). The straw sticks up an inch above the top. About how much work does it take to suck up the milkshake through the straw (neglecting friction)? Answer in inch-ounces.

Short Answer

Expert verified
Calculate the volume using the truncated cone formula, find the weight by multiplying with density, determine the center of mass, and apply the work formula.

Step by step solution

01

Find the Volume of the Truncated Cone

The volume of a truncated cone can be calculated using the formula: \[ V = \frac{1}{3} \pi h ( R^2 + Rr + r^2 ) \] where \( h \) is the height (7 inches), \( R \) is the radius of the top (1.75 inches), and \( r \) is the radius of the base (1.25 inches).
02

Calculate the Volume

Substituting the values into the formula: \[V = \frac{1}{3} \pi \times 7 \times (1.75^2 + 1.75 \times 1.25 + 1.25^2).\] Calculate the expression to find \( V \).
03

Compute the Weight of the Milkshake

The density of the milkshake is given as \( \frac{4}{9} \) ounces per cubic inch. Multiply this by the volume calculated in the previous step to find the total weight of the milkshake: \( \text{Weight} = \frac{4}{9} \times V \).
04

Determine the Distance the Milkshake Moves

The milkshake has to be lifted from a height of the center of mass of the truncated cone to the top of the straw, which is 1 inch above the top of the container. The distance lifted is the distance from the center of mass to the top plus 1 inch.
05

Calculate the Position of the Center of Mass

For a truncated cone, the center of mass can be found using proportional divisions. The formula is: \[ \bar{z} = \frac{h}{4} \cdot \frac{R^2 + 2Rr + 3r^2}{R^2 + Rr + r^2} \] Substitute \( h = 7 \), \( R = 1.75 \), and \( r = 1.25 \) to find the height of the center of mass from the base.
06

Calculate the Work Done

Work is calculated as \( \text{Work} = \text{Weight} \times \text{Distance} \). Use the weight from Step 3 and the total distance from the center of mass to the top of the straw from Step 4.

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Key Concepts

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

Truncated Cone Volume
A truncated cone, also known as a frustum, is a cone with its top cut off parallel to its base. Calculating its volume is essential when you deal with objects like drinking cups or certain containers.
To find the volume of a truncated cone, you use the formula:
  • \[ V = \frac{1}{3} \pi h ( R^2 + Rr + r^2 ) \]
In this formula:
  • \( V \) is the volume.
  • \( h \) is the height of the truncated cone.
  • \( R \) is the radius of the larger (top) circle.
  • \( r \) is the radius of the smaller (bottom) circle.
In our problem, the height \( h \) is 7 inches, the top radius \( R \) is 1.75 inches (half of 3.5 inches), and the bottom radius \( r \) is 1.25 inches (half of 2.5 inches).
By substituting these values into the formula, we can compute the volume of our milkshake container.
Density Calculations
Density is a measure of how much mass is contained in a given volume. It is calculated by dividing mass by volume and often expressed in units such as ounces per cubic inch in the case of our milkshake.
To calculate the total weight of the milkshake, we need to multiply the density by the volume obtained earlier. The density of the milkshake given is \( \frac{4}{9} \) ounces per cubic inch.
This step simply involves taking the volume \( V \) that you calculated in the previous section and applying this density formula:
  • \[ \text{Weight} = \text{Density} \, \times \, \text{Volume} \]
By substituting the calculated volume for \( V \), you can find the total weight of the milkshake, which will be used in further calculations.
Center of Mass
Finding the center of mass of a truncated cone involves understanding how mass is distributed along its height. This point is crucial when calculating work, as it affects how far you need to lift the milkshake.
For a truncated cone, the center of mass can be located using:
  • \[ \bar{z} = \frac{h}{4} \cdot \frac{R^2 + 2Rr + 3r^2}{R^2 + Rr + r^2} \]
This formula involves:
  • \( h \), the height of the truncated cone.
  • \( R \) and \( r \), the radii of the larger and smaller bases respectively.
By substituting the given values of \( h = 7 \), \( R = 1.75 \), and \( r = 1.25 \), you can pinpoint where the center of mass is located. This midpoint helps determine how far the milkshake must be lifted.
Work-Energy Principle
The work-energy principle states that work done on an object is equal to the change in its energy. When you suck a milkshake through a straw, you are performing work on it by lifting it against gravity from the container to your mouth.
The work done is calculated using:
  • \[ \text{Work} = \text{Weight} \times \text{Distance} \]
In this case:
  • "Weight" is the total weight of the milkshake as calculated previously.
  • "Distance" is the path from the center of mass to the top of the straw, accounting for the additional 1-inch length above the container.
By multiplying these values, you obtain the total work done in moving the milkshake through the straw. This provides an understanding of the energy required in lifting the entire content of the container against gravity.

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Most popular questions from this chapter

Find the volume of the solid generated by revolving the region bounded by the parabola \(y=x^{2}\) and the line \(y=1\) about a. the line \(y=1 . \quad\) b. the line \(y=2\) c. the line \(y=-1\)

An alternative derivation of the surface area formula Assume \(f\) is smooth on \([a, b]\) and partition \([a, b]\) in the usual way. In the \(k\) th subinterval \(\left[x_{k-1}, x_{k}\right]\) construct the tangent line to the curve at the midpoint \(m_{k}=\left(x_{k-1}+x_{k}\right) / 2,\) as in the figure here. a. Show that \(r_{1}=f\left(m_{k}\right)-f^{\prime}\left(m_{k}\right) \frac{\Delta x_{k}}{2}\) and \(r_{2}=f\left(m_{k}\right)+\) \(\quad f^{\prime}\left(m_{k}\right) \frac{\Delta x_{k}}{2}\) b. Show that the length \(L_{k}\) of the tangent line segment in the \(k\) th subinterval is \(L_{k}=\sqrt{\left(\Delta x_{k}\right)^{2}+\left(f^{\prime}\left(m_{k}\right) \Delta x_{k}\right)^{2}}\) Graph cannot copy c. Show that the lateral surface area of the frustum of the cone swept out by the tangent line segment as it revolves about the \(x\) -axis is 2\(\pi f\left(m_{k}\right) \sqrt{1+\left(f^{\prime}\left(m_{k}\right)\right)^{2}} \Delta x_{k}\) d. Show that the area of the surface generated by revolving \(y=f(x)\) about the \(x\) -axis over \([a, b]\) is $$ \lim _{n \rightarrow \infty} \sum_{k=1}^{n}\left(\begin{array}{l}{\text { lateral surface area }} \\ {\text { of } k \text { th frustum }}\end{array}\right)=\int_{a}^{b} 2 \pi f(x) \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x $$

The region in the first quadrant that is bounded above by the curve \(y=1 / \sqrt{x},\) on the left by the line \(x=1 / 4,\) and below by the line \(y=1\) is revolved about the \(y\) -axis to generate a solid. Find the volume of the solid by a. the washer method. b. the shell method.

Use the shell method to find the volumes of the solids generated by re- volving the regions bounded by the curves and lines about the \(y\)-axis. \(y=x^{2}, \quad y=2-x, \quad x=0,\) for \(x \geq 0\)

In Exercises \(31-36,\) use a CAS to perform the following steps for the given curve over the closed interval. a. Plot the curve together with the polygonal path approximations for \(n=2,4,8\) partition points over the interval. (See Figure \(6.24 . )\) b. Find the corresponding approximation to the length of the curve by summing the lengths of the line segments. c. Evaluate the length of the curve using an integral. Compare your appoximations for \(n=2,4,8\) with the actual compare= given by the integral. How does the actual length compare with the approximations as \(n\) increases? Explain your answer. $$ f(x)=x^{1 / 3}+x^{2 / 3}, \quad 0 \leq x \leq 2 $$
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