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You are designing a part for a piece of machinery. The part consists of a piece of sheet metal cut as shown below. The shape of the upper edge of the part is given by y/₁(a), and the shape of the lower edge of the part is given by y/2(a). 31 (2)=h(z) * 3/2(x)=h()* where h 6.6 m and d = 3.4 m y₁(x) h- 12EA dm -X₂(x) dx d You decide to find the moment of inertia of the part about that y axis first. The mass density per area for the sheet metal is 3 kg/m^2. In order to find the moment of inertia, first you must chop the part into small mass elements, dm's, that you know the moments of inertial for, dl's. Then you must use an integral to sum up all of the di's.

You are designing a part for a piece of machinery. The part consists of a piece of sheet metal cut as shown below. The shape of the upper edge of the part is given by y/₁(a), and the shape of the lower edge of the part is given by y/2(a). 31 (2)=h(z) * 3/2(x)=h()* where h 6.6 m and d = 3.4 m y₁(x) h- 12EA dm -X₂(x) dx d You decide to find the moment of inertia of the part about that y axis first. The mass density per area for the sheet metal is 3 kg/m^2. In order to find the moment of inertia, first you must chop the part into small mass elements, dm's, that you know the moments of inertial for, dl's. Then you must use an integral to sum up all of the di's.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Need help with this last question in this question set. Any help will be apprecaited. I tried to fit all the nessicary items into a clear screenshot. Thanks 

Distance to dm Differential Area Differential Mass Differential Moment Moment of Inertia (y) Now find the total moment of inertia by adding up (integrating) all of the dl's. To do this, put everything in terms of x and compute the integral over x. Don't forget about limits. Then plug in the values for your constants so that you can get a value for the moment of inertia. Enter the numerical value for your Iy below. A Hint About the Limits of Integration A Hint About Your Integral Iy = Total Mass 128.28kg*m^2 Find the total mass of the part by adding up (integrating) all of the dm's. To do this, put everything in terms of x and compute the integral over x. Don't forget about limits. Then plug in the values for your constants so that you can get a numerical value for the total mass. m = 40.392kg Center of Mass (x) Find the x position of the center of mass of the part. You may need to look up the definiti for the center of mass of a continuous object if you do not remember it. Here are the steps that are involved in the calculation of the center of mass. Break the object into a bunch of little dm's like you did for the moment of inertia. Multiply the dm by the x coordinate of its center of mass. Let's call this x cm,dm. Add up all of the dm, integrate x cm,dm. Divide by the total mass. x cm = 1.574m Moment of Inertia (x) Iz くくく Now you will find the total moment of inertia about the x axis. There are at least two ways to find this. You can try and find it by using an integral with respect to x. This way turns out to by much more difficult than setting up an integral expression with respect to y instead. If you set up this integral with respect to y, you should notice that the expression for the moment of inertia about the x axis is very similar to the expression for the moment of inertia about the y axis. If you think about this in the correct way, then you can actually figure this integral out with the work you have already done for Iy and not have to integrate again. Remember that the integration variable is arbitrary for a definite integral, as it is summed over when calculating the integral. So you are allowed to make the replacement y → x. Also, if you make the replacements h d and d→ h, then the expression for I would be exactly the same as that for Iy. So you should be able to make these replacements in your final equation for Iy to give you the final equation for I. Enter the numerical value for your I below. = 128kg*m^2
Transcribed Image Text:Distance to dm Differential Area Differential Mass Differential Moment Moment of Inertia (y) Now find the total moment of inertia by adding up (integrating) all of the dl's. To do this, put everything in terms of x and compute the integral over x. Don't forget about limits. Then plug in the values for your constants so that you can get a value for the moment of inertia. Enter the numerical value for your Iy below. A Hint About the Limits of Integration A Hint About Your Integral Iy = Total Mass 128.28kg*m^2 Find the total mass of the part by adding up (integrating) all of the dm's. To do this, put everything in terms of x and compute the integral over x. Don't forget about limits. Then plug in the values for your constants so that you can get a numerical value for the total mass. m = 40.392kg Center of Mass (x) Find the x position of the center of mass of the part. You may need to look up the definiti for the center of mass of a continuous object if you do not remember it. Here are the steps that are involved in the calculation of the center of mass. Break the object into a bunch of little dm's like you did for the moment of inertia. Multiply the dm by the x coordinate of its center of mass. Let's call this x cm,dm. Add up all of the dm, integrate x cm,dm. Divide by the total mass. x cm = 1.574m Moment of Inertia (x) Iz くくく Now you will find the total moment of inertia about the x axis. There are at least two ways to find this. You can try and find it by using an integral with respect to x. This way turns out to by much more difficult than setting up an integral expression with respect to y instead. If you set up this integral with respect to y, you should notice that the expression for the moment of inertia about the x axis is very similar to the expression for the moment of inertia about the y axis. If you think about this in the correct way, then you can actually figure this integral out with the work you have already done for Iy and not have to integrate again. Remember that the integration variable is arbitrary for a definite integral, as it is summed over when calculating the integral. So you are allowed to make the replacement y → x. Also, if you make the replacements h d and d→ h, then the expression for I would be exactly the same as that for Iy. So you should be able to make these replacements in your final equation for Iy to give you the final equation for I. Enter the numerical value for your I below. = 128kg*m^2
You are designing a part for a piece of machinery. The part consists of a piece of sheet metal cut as shown below. The shape of the upper edge of the part is given by y₁(x), and the shape of the lower edge of the part is given by y2(x). ¥₁(x) = h ( z ) ² 4 y2(x) = h ( z ) ¹ where h = 6.6 m and d = 3.4 m y, (x) dx •½/₂(x) X You decide to find the moment of inertia of the part about that y axis first. The mass density per area for the sheet metal is 3 kg/m^2. In order to find the moment of inertia, first you must chop the part into small mass elements, dm's, that you know the moments of inertia for, dl's. Then you must use an integral to sum up all of the dl's.
Transcribed Image Text:You are designing a part for a piece of machinery. The part consists of a piece of sheet metal cut as shown below. The shape of the upper edge of the part is given by y₁(x), and the shape of the lower edge of the part is given by y2(x). ¥₁(x) = h ( z ) ² 4 y2(x) = h ( z ) ¹ where h = 6.6 m and d = 3.4 m y, (x) dx •½/₂(x) X You decide to find the moment of inertia of the part about that y axis first. The mass density per area for the sheet metal is 3 kg/m^2. In order to find the moment of inertia, first you must chop the part into small mass elements, dm's, that you know the moments of inertia for, dl's. Then you must use an integral to sum up all of the dl's.
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