The moment of inertia tensor is symmetric, and is related to the angular momentum vector by. \nonumber \]. moment of inertia in kg*m2. In the case with the axis in the center of the barbell, each of the two masses m is a distance \(R\) away from the axis, giving a moment of inertia of, \[I_{1} = mR^{2} + mR^{2} = 2mR^{2} \ldotp\], In the case with the axis at the end of the barbellpassing through one of the massesthe moment of inertia is, \[I_{2} = m(0)^{2} + m(2R)^{2} = 4mR^{2} \ldotp\]. horizontal strips when you want to find the moment of inertia about the \(x\) axis and vertical strips for the moment of inertia about the \(y\) axis. }\tag{10.2.8} \end{align}, \begin{align} J_O \amp = \int_0^r \rho^2\ 2\pi\rho \ d\rho\notag\\ \amp = 2 \pi \int_0^r \rho^3 d\rho\notag\\ \amp = 2 \pi \left [ \frac{\rho^4}{4}\right ]_0^r\notag\\ J_O \amp = \frac{\pi r^4}{2}\text{. Trebuchets can launch objects from 500 to 1,000 feet. }\), Following the same procedure as before, we divide the rectangle into square differential elements \(dA = dx\ dy\) and evaluate the double integral for \(I_y\) from (10.1.3) first by integrating over \(x\text{,}\) and then over \(y\text{. mm 4; cm 4; m 4; Converting between Units. This approach is illustrated in the next example. The convention is to place a bar over the symbol \(I\) when the the axis is centroidal. A moving body keeps moving not because of its inertia but only because of the absence of a . }\) Note that the \(y^2\) term can be taken out of the inside integral, because in terms of \(x\text{,}\) it is constant. It is an extensive (additive) property: the moment of . Click Content tabCalculation panelMoment of Inertia. The moment of inertia is defined as the quantity reflected by the body resisting angular acceleration, which is the sum of the product of each particle's mass and its square of the distance from the axis of rotation. A long arm is attached to fulcrum, with one short (significantly shorter) arm attached to a heavy counterbalance and a long arm with a sling attached. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Applying our previous result (10.2.2) to a vertical strip with height \(h\) and infinitesimal width \(dx\) gives the strip's differential moment of inertia. Weak axis: I z = 20 m m ( 200 m m) 3 12 + ( 200 m m 20 m m 10 m m) ( 10 m m) 3 12 + 10 m m ( 100 m m) 3 12 = 1.418 10 7 m m 4. \begin{equation} I_x = \frac{bh^3}{12}\label{MOI-triangle-base}\tag{10.2.4} \end{equation}, As we did when finding centroids in Section 7.7 we need to evaluate the bounding function of the triangle. (Moment of inertia)(Rotational acceleration) omega2= omegao2+2(rotational acceleration)(0) 77. Review. The differential element dA has width dx and height dy, so dA = dx dy = dy dx. The moment of inertia about the vertical centerline is the same. In physics and applied mathematics, the mass moment of inertia, usually denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass.Mass moments of inertia have units of dimension ML 2 ([mass] [length] 2).It should not be confused with the second moment of area, which is used in beam calculations. From this result, we can conclude that it is twice as hard to rotate the barbell about the end than about its center. }\tag{10.2.11} \end{equation}, Similarly, the moment of inertia of a quarter circle is half the moment of inertia of a semi-circle, so, \begin{equation} I_x = I_y = \frac{\pi r^4}{16}\text{. Doubling the width of the rectangle will double \(I_x\) but doubling the height will increase \(I_x\) eightfold. This is the same result that we saw previously (10.2.3) after integrating the inside integral for the moment of inertia of a rectangle. For vertical strips, which are parallel to the \(y\) axis we can use the definition of the Moment of Inertia. The integration techniques demonstrated can be used to find the moment of inertia of any two-dimensional shape about any desired axis. The higher the moment of inertia, the more resistant a body is to angular rotation. A pendulum in the shape of a rod (Figure \(\PageIndex{8}\)) is released from rest at an angle of 30. When using strips which are parallel to the axis of interest is impractical mathematically, the alternative is to use strips which are perpendicular to the axis. In its inertial properties, the body behaves like a circular cylinder. Letting \(dA = y\ dx\) and substituting \(y = f(x) = x^3 +x\) we have, \begin{align*} I_y \amp = \int_A x^2\ dA\\ \amp = \int_0^1 x^2 y\ dx\\ \amp = \int_0^1 x^2 (x^3+x)\ dx\\ \amp = \int_0^1 (x^5 + x^3) dx\\ \amp = \left . \[U = mgh_{cm} = mgL^2 (\cos \theta). We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. The formula for \(I_y\) is the same as the formula as we found previously for \(I_x\) except that the base and height terms have reversed roles. Internal forces in a beam caused by an external load. where I is the moment of inertia of the throwing arm. Noting that the polar moment of inertia of a shape is the sum of its rectangular moments of inertia and that \(I_x\) and \(I_y\) are equal for a circle due to its symmetry. This is the moment of inertia of a right triangle about an axis passing through its base. We saw in the last section that when solving (10.1.3) the double integration could be conducted in either order, and that the result of completing the inside integral was a single integral. In rotational motion, moment of inertia is extremely important as a variety of questions can be framed from this topic. (Bookshelves/Mechanical_Engineering/Engineering_Statics:_Open_and_Interactive_(Baker_and_Haynes)/10:_Moments_of_Inertia/10.02:_Moments_of_Inertia_of_Common_Shapes), /content/body/div[4]/article/div/dl[2]/dd/p[9]/span, line 1, column 6, Moment of Inertia of a Differential Strip, Circles, Semicircles, and Quarter-circles, status page at https://status.libretexts.org. The rod has length 0.5 m and mass 2.0 kg. The moment of inertia of a point mass is given by I = mr 2, but the rod would have to be considered to be an infinite number of point masses, and each must be multiplied by the square of its distance . The Trechbuchet works entirely on gravitational potential energy. Example 10.4.1. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. This is why the arm is tapered on many trebuchets. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Every rigid object has a de nite moment of inertia about a particular axis of rotation. That is, a body with high moment of inertia resists angular acceleration, so if it is not . We again start with the relationship for the surface mass density, which is the mass per unit surface area. The radius of the sphere is 20.0 cm and has mass 1.0 kg. 77 two blocks are connected by a string of negligible mass passing over a pulley of radius r = 0. Find Select the object to which you want to calculate the moment of inertia, and press Enter. It is based not only on the physical shape of the object and its distribution of mass but also the specific configuration of how the object is rotating. This is the polar moment of inertia of a circle about a point at its center. Consider a uniform (density and shape) thin rod of mass M and length L as shown in Figure \(\PageIndex{3}\). Moment of inertia also known as the angular mass or rotational inertia can be defined w.r.t. This radius range then becomes our limits of integration for \(dr\), that is, we integrate from \(r = 0\) to \(r = R\). Legal. A beam with more material farther from the neutral axis will have a larger moment of inertia and be stiffer. If you would like to avoid double integration, you may use vertical or horizontal strips, but you must take care to apply the correct integral. In most cases, \(h\) will be a function of \(x\text{. Find the moment of inertia of the rectangle about the \(y\) axis using square differential elements (dA\text{. Use conservation of energy to solve the problem. Mechanics of a Simple Trebuchet Mechanics of a Simple Trebuchet Also Define M = Mass of the Beam (m1 + m2) L = Length of the Beam (l1 + l2) Torque Moment of Inertia Define Numerical Approximation: These functions can be used to determine q and w after a time Dt. This moment at a point on the face increases with with the square of the distance \(y\) of the point from the neutral axis because both the internal force and the moment arm are proportional to this distance. The expression for \(dI_x\) assumes that the vertical strip has a lower bound on the \(x\) axis. The appearance of \(y^2\) in this relationship is what connects a bending beam to the area moment of inertia. Notice that the centroidal moment of inertia of the rectangle is smaller than the corresponding moment of inertia about the baseline. We therefore need to find a way to relate mass to spatial variables. \end{align*}, Similarly we will find \(I_x\) using horizontal strips, by evaluating this integral with \(dA = (b-x) dy\), \begin{align*} I_x \amp = \int_A y^2 dA \text{.} In (a), the center of mass of the sphere is located at a distance \(L + R\) from the axis of rotation. For a uniform solid triaxial ellipsoid, the moments of inertia are A = 1 5m(b2 + c2) B = 1 5m(c2 + a2) C = 1 5m(c2 + a2) The most straightforward approach is to use the definitions of the moment of inertia (10.1.3) along with strips parallel to the designated axis, i.e. The name for I is moment of inertia. \begin{equation} I_x = \bar{I}_y = \frac{\pi r^4}{8}\text{. If you are new to structural design, then check out our design tutorials where you can learn how to use the moment of inertia to design structural elements such as. The moment of inertia, otherwise known as the angular mass or rotational inertia, of a rigid body is a tensor that determines the torque needed for a desired angular acceleration about a rotational axis. The trebuchet, mistaken most commonly as a catapult, is an ancient weapon used primarily by Norsemen in the Middle Ages. Here, the horizontal dimension is cubed and the vertical dimension is the linear term. The moment of inertia of an element of mass located a distance from the center of rotation is. The moment of inertia of a point mass with respect to an axis is defined as the product of the mass times the distance from the axis squared. We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. Now consider a compound object such as that in Figure \(\PageIndex{6}\), which depicts a thin disk at the end of a thin rod. Luckily there is an easier way to go about it. Then evaluate the differential equation numerically. In this section, we will use polar coordinates and symmetry to find the moments of inertia of circles, semi-circles and quarter-circles. We see that the moment of inertia is greater in (a) than (b). At the bottom of the swing, K = \(\frac{1}{2} I \omega^{2}\). This rectangle is oriented with its bottom-left corner at the origin and its upper-right corner at the point \((b,h)\text{,}\) where \(b\) and \(h\) are constants. When an elastic beam is loaded from above, it will sag. It represents the rotational inertia of an object. Moment of inertia comes under the chapter of rotational motion in mechanics. It is only constant for a particular rigid body and a particular axis of rotation. inches 4; Area Moment of Inertia - Metric units. It is best to work out specific examples in detail to get a feel for how to calculate the moment of inertia for specific shapes. As discussed in Subsection 10.1.3, a moment of inertia about an axis passing through the area's centroid is a Centroidal Moment of Inertia. Because r is the distance to the axis of rotation from each piece of mass that makes up the object, the moment of inertia for any object depends on the chosen axis. Inertia is a passive property and does not enable a body to do anything except oppose such active agents as forces and torques. }\label{Ix-rectangle}\tag{10.2.2} \end{equation}. This will allow us to set up a problem as a single integral using strips and skip the inside integral completely as we will see in Subsection 10.2.2. Enter a text for the description of the moment of inertia block. Therefore: \[\Delta U + \Delta K = 0 \Rightarrow (mg \frac{L}{2} (1 - \cos \theta) - 0) + (0 - \frac{1}{2} I \omega^{2}) = 0 \nonumber\], \[\frac{1}{2} I \omega^{2} = mg \frac{L}{2} (1 - \cos \theta) \ldotp \nonumber\], \[\omega = \sqrt{mg \frac{L}{I} (1 - \cos \theta)} = \sqrt{mg \frac{L}{\frac{1}{3} mL^{2}} (1 - \cos \theta)} = \sqrt{g \frac{3}{L} (1 - \cos \theta)} \ldotp \nonumber\], \[\omega = \sqrt{(9.8\; m/s^{2}) \left(\dfrac{3}{0.3\; m}\right) (1 - \cos 30)} = 3.6\; rad/s \ldotp \nonumber\]. This happens because more mass is distributed farther from the axis of rotation. The moments of inertia of a mass have units of dimension ML 2 ( [mass] [length] 2 ). The potential . The internal forces sum to zero in the horizontal direction, but they produce a net couple-moment which resists the external bending moment. Assume that some external load is causing an external bending moment which is opposed by the internal forces exposed at a cut. Moment of Inertia Example 2: FLYWHEEL of an automobile. Moment of inertia can be defined as the quantitative measure of a body's rotational inertia.Simply put, the moment of inertia can be described as a quantity that decides the amount of torque needed for a specific angular acceleration in a rotational axis. rotation axis, as a quantity that decides the amount of torque required for a desired angular acceleration or a property of a body due to which it resists angular acceleration. It would seem like this is an insignificant difference, but the order of \(dx\) and \(dy\) in this expression determines the order of integration of the double integral. Any idea what the moment of inertia in J in kg.m2 is please? This means when the rigidbody moves and rotates in space, the moment of inertia in worldspace keeps aligned with the worldspace axis of the body. What is the moment of inertia of a cylinder of radius \(R\) and mass \(m\) about an axis through a point on the surface, as shown below? In these diagrams, the centroidal axes are red, and moments of inertia about centroidal axes are indicated by the overbar. Since the mass and size of the child are much smaller than the merry-go-round, we can approximate the child as a point mass. We will begin with the simplest case: the moment of inertia of a rectangle about a horizontal axis located at its base. }\label{straight-line}\tag{10.2.5} \end{equation}, By inspection we see that the a vertical strip extends from the \(x\) axis to the function so \(dA= y\ dx\text{. ! This case arises frequently and is especially simple because the boundaries of the shape are all constants. The notation we use is mc = 25 kg, rc = 1.0 m, mm = 500 kg, rm = 2.0 m. Our goal is to find \(I_{total} = \sum_{i} I_{i}\) (Equation \ref{10.21}). Date Final Exam MEEN 225, Engineering Mechanics PROBLEM #1 (20 points) Two blocks A and B have a weight of 10 lb and 6 The principal moments of inertia are given by the entries in the diagonalized moment of inertia matrix . Use the fact that moments of inertia simply add, namely Itotal = I1 + I2 + I3 + , where I1 is the moment of inertia of the object you want to measure and I2, I3, are the moments of We will use these observations to optimize the process of finding moments of inertia for other shapes by avoiding double integration. Now we use a simplification for the area. Moment of Inertia is a very useful term for mechanical engineering and piping stress analysis. The payload could be thrown a far distance and do considerable damage, either by smashing down walls or striking the enemy while inside their stronghold. \left( \frac{x^4}{16} - \frac{x^5}{12} \right )\right \vert_0^{1/2}\\ \amp= \left( \frac{({1/2})^4}{16} - \frac, For vertical strips, which are perpendicular to the \(x\) axis, we will take subtract the moment of inertia of the area below \(y_1\) from the moment of inertia of the area below \(y_2\text{. In (b), the center of mass of the sphere is located a distance \(R\) from the axis of rotation. This result makes it much easier to find \(I_x\) for the spandrel that was nearly impossible to find with horizontal strips. In this subsection, we show how to calculate the moment of inertia for several standard types of objects, as well as how to use known moments of inertia to find the moment of inertia for a shifted axis or for a compound object. What is the moment of inertia of this rectangle with respect to the \(x\) axis? The moment of inertia of any extended object is built up from that basic definition. \[\begin{split} I_{total} & = \sum_{i} I_{i} = I_{Rod} + I_{Sphere}; \\ I_{Sphere} & = I_{center\; of\; mass} + m_{Sphere} (L + R)^{2} = \frac{2}{5} m_{Sphere} R^{2} + m_{Sphere} (L + R)^{2}; \\ I_{total} & = I_{Rod} + I_{Sphere} = \frac{1}{3} m_{Rod} L^{2} + \frac{2}{5} m_{Sphere} R^{2} + m_{Sphere} (L + R)^{2}; \\ I_{total} & = \frac{1}{3} (20\; kg)(0.5\; m)^{2} + \frac{2}{5} (1.0\; kg)(0.2\; m)^{2} + (1.0\; kg)(0.5\; m + 0.2\; m)^{2}; \\ I_{total} & = (0.167 + 0.016 + 0.490)\; kg\; \cdotp m^{2} = 0.673\; kg\; \cdotp m^{2} \ldotp \end{split}\], \[\begin{split} I_{Sphere} & = \frac{2}{5} m_{Sphere} R^{2} + m_{Sphere} R^{2}; \\ I_{total} & = I_{Rod} + I_{Sphere} = \frac{1}{3} m_{Rod} L^{2} + \frac{2}{5} (1.0\; kg)(0.2\; m)^{2} + (1.0\; kg)(0.2\; m)^{2}; \\ I_{total} & = (0.167 + 0.016 + 0.04)\; kg\; \cdotp m^{2} = 0.223\; kg\; \cdotp m^{2} \ldotp \end{split}\]. You have three 24 ft long wooden 2 6's and you want to nail them together them to make the stiffest possible beam. \frac{x^3}{3} \right |_0^b \\ I_y \amp = \frac{hb^3}{3} \end{align*}. This solution demonstrates that the result is the same when the order of integration is reversed. \nonumber \]. It actually is just a property of a shape and is used in the analysis of how some Here is a summary of the alternate approaches to finding the moment of inertia of a shape using integration. moment of inertia, in physics, quantitative measure of the rotational inertia of a bodyi.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). Symbolically, this unit of measurement is kg-m2. The moment of inertia of a region can be computed in the Wolfram Language using MomentOfInertia [ reg ]. Thanks in advance. Have tried the manufacturer but it's like trying to pull chicken teeth! The moment of inertia is not an intrinsic property of the body, but rather depends on the choice of the point around which the body rotates. For best performance, the moment of inertia of the arm should be as small as possible. The rod extends from \(x = 0\) to \(x = L\), since the axis is at the end of the rod at \(x = 0\). }\label{dI_y}\tag{10.2.7} \end{align}, The width \(b\) will usually have to be expressed as a function of \(y\text{.}\). Legal. I parallel-axis = 1 2 m d R 2 + m d ( L + R) 2. When used in an equation, the moment of . It is important to note that the moments of inertia of the objects in Equation \(\PageIndex{6}\) are about a common axis. The moment of inertia can be derived as getting the moment of inertia of the parts and applying the transfer formula: I = I 0 + Ad 2. In this article, we will explore more about the Moment of Inertia, Its definition, formulas, units, equations, and applications. (A.19) In general, when an object is in angular motion, the mass elements in the body are located at different distances from the center of . To take advantage of the geometry of a circle, we'll divide the area into thin rings, as shown in the diagram, and define the distance from the origin to a point on the ring as \(\rho\text{. }\) The reason for using thin rings for \(dA\) is the same reason we used strips parallel to the axis of interest to find \(I_x\) and \(I_y\text{;}\) all points on the differential ring are the same distance from the origin, so we can find the moment of inertia using single integration. Recall that in our derivation of this equation, each piece of mass had the same magnitude of velocity, which means the whole piece had to have a single distance r to the axis of rotation. Pay attention to the placement of the axis with respect to the shape, because if the axis is located elsewhere or oriented differently, the results will be different. We defined the moment of inertia I of an object to be (10.6.1) I = i m i r i 2 for all the point masses that make up the object. This is the focus of most of the rest of this section. Since it is uniform, the surface mass density \(\sigma\) is constant: \[\sigma = \frac{m}{A}\] or \[\sigma A = m\] so \[dm = \sigma (dA)\]. The moment of inertia, otherwise known as the angular mass or rotational inertia, of a rigid body is a tensor that determines the torque needed for a desired angular acceleration about a rotational axis. The horizontal distance the payload would travel is called the trebuchet's range. for all the point masses that make up the object. Because \(r\) is the distance to the axis of rotation from each piece of mass that makes up the object, the moment of inertia for any object depends on the chosen axis. This page titled 10.2: Moments of Inertia of Common Shapes is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Daniel W. Baker and William Haynes (Engineeringstatics) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. (5), the moment of inertia depends on the axis of rotation. }\), \begin{align} I_x \amp= \frac{bh^3}{3} \amp \amp \rightarrow \amp dI_x \amp= \frac{h^3}{3} dx\text{. This is the moment of inertia of a circle about a vertical or horizontal axis passing through its center. The axis may be internal or external and may or may not be fixed. This is the formula for the moment of inertia of a rectangle about an axis passing through its base, and is worth remembering. The moment of inertia formula is important for students. The simple analogy is that of a rod. In following sections we will use the integral definitions of moment of inertia (10.1.3) to find the moments of inertia of five common shapes: rectangle, triangle, circle, semi-circle, and quarter-circle with respect to a specified axis. In the case of this object, that would be a rod of length L rotating about its end, and a thin disk of radius \(R\) rotating about an axis shifted off of the center by a distance \(L + R\), where \(R\) is the radius of the disk. Lecture 11: Mass Moment of Inertia of Rigid Bodies Viewing videos requires an internet connection Description: Prof. Vandiver goes over the definition of the moment of inertia matrix, principle axes and symmetry rules, example computation of Izz for a disk, and the parallel axis theorem. \[ I_y = \frac{hb^3}{12} \text{.} The similarity between the process of finding the moment of inertia of a rod about an axis through its middle and about an axis through its end is striking, and suggests that there might be a simpler method for determining the moment of inertia for a rod about any axis parallel to the axis through the center of mass. Once this has been done, evaluating the integral is straightforward. The trebuchet was preferred over a catapult due to its greater range capability and greater accuracy. Now, we will evaluate (10.1.3) using \(dA = dy\ dx\) which reverses the order of integration and means that the integral over \(y\) gets conducted first. Here are a couple of examples of the expression for I for two special objects: The general form of the moment of inertia involves an integral. First, we will evaluate (10.1.3) using \(dA = dx\ dy\text{. We have found that the moment of inertia of a rectangle about an axis through its base is (10.2.2), the same as before. If you use vertical strips to find \(I_y\) or horizontal strips to find \(I_x\text{,}\) then you can still use (10.1.3), but skip the double integration. The mass moment of inertia depends on the distribution of . Clearly, a better approach would be helpful. The moment of inertia of an element of mass located a distance from the center of rotation is. A list of formulas for the moment of inertia of different shapes can be found here. However, if we go back to the initial definition of moment of inertia as a summation, we can reason that a compound objects moment of inertia can be found from the sum of each part of the object: \[I_{total} = \sum_{i} I_{i} \ldotp \label{10.21}\]. One of the most advanced siege engines used in the Middle Ages was the trebuchet, which used a large counterweight to store energy to launch a payload, or projectile. We are expressing \(dA\) in terms of \(dy\text{,}\) so everything inside the integral must be constant or expressed in terms of \(y\) in order to integrate. Moment of Inertia behaves as angular mass and is called rotational inertia. }\), \begin{align*} I_y \amp = \int_A x^2 dA \\ \amp = \int_0^h \int_0^b x^2\ dx\ dy\\ \amp = \int_0^h \left [ \int_0^b x^2\ dx \right ] \ dy\\ \amp = \int_0^h \left [ \frac{x^3}{3}\right ]_0^b \ dy\\ \amp = \int_0^h \boxed{\frac{b^3}{3} dy} \\ \amp = \frac{b^3}{3} y \Big |_0^h \\ I_y \amp = \frac{b^3h}{3} \end{align*}. It has a length 30 cm and mass 300 g. What is its angular velocity at its lowest point? View Practice Exam 3.pdf from MEEN 225 at Texas A&M University. In both cases, the moment of inertia of the rod is about an axis at one end. Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. \begin{align*} I_y \amp = \int x^2 dA\\ \amp = \int_0^{0.5} {x^2} \left ( \frac{x}{4} - \frac{x^2}{2} \right ) dx\\ \amp= \int_0^{1/2} \left( \frac{x^3}{4} - \frac{x^4}{2} \right) dx \\ \amp= \left . homework-and-exercises newtonian-mechanics rotational-dynamics torque moment-of-inertia Share Cite Improve this question Follow The strip must be parallel in order for (10.1.3) to work; when parallel, all parts of the strip are the same distance from the axis. \end{align*}, \begin{align*} I_x \amp = 3.49 \times \cm{10^{-6}}^4 \amp I_y \amp = 7.81 \times \cm{10^{-6}}^4 \end{align*}, \begin{align*} y_2 \amp = x/4 \amp y_2 \amp = x^2/2 \end{align*}, By equating the two functions, we learn that they intersect at \((0,0)\) and \((1/2,1/8)\text{,}\) so the limits on \(x\) are \(x=0\) and \(x=1/2\text{. 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