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Do Chapter 12, Problem 5. This is an integration problem, to calculate the center of mass (center of gravity) for a continuous distribution. Specifically, we need the horizontal component of the center of mass. The height varies from h to zero according to this function: y(x) = h ( – 1)*. The constants h and e replace 1.00 m and 3.00 m. There is also a thickness t and a density p. You need two integrals, the total mass and the center of mass. Possibly surprisingly, you don't actually need the numbers t, h, and p.

Do Chapter 12, Problem 5. This is an integration problem, to calculate the center of mass (center of gravity) for a continuous distribution. Specifically, we need the horizontal component of the center of mass. The height varies from h to zero according to this function: y(x) = h ( – 1)*. The constants h and e replace 1.00 m and 3.00 m. There is also a thickness t and a density p. You need two integrals, the total mass and the center of mass. Possibly surprisingly, you don't actually need the numbers t, h, and p.

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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Do Chapter 12, Problem 5.  This is an integration problem, to calculate the center of mass (center of gravity) for a continuous distribution.  Specifically, we need the horizontal component of the center of mass.

The height varies from h to zero according to this function: y(x)=h(xℓ−1)2y(x)=h(xℓ−1)2 .  The constants h and ℓ replace 1.00 m and 3.00 m.  There is also a thickness t and a density ρ.  You need two integrals, the total mass and the center of mass.  Possibly surprisingly, you don't actually need the numbers t, h, and ρ.

Do Chapter 12, Problem 5. This is an integration problem, to calculate the center of mass (center of gravity) for a continuous distribution. Specifically, we need the horizontal component of the center of mass. The height varies from h to zero according to this function: y(x) = h ( - 1). The constants h and e replace 1.00 m and 3.00 m. There is also a thickness t and a density p. You need two integrals, the total mass and the center of mass. Possibly surprisingly, you don't actually need the numbers t, h, and p. Ax y(x) The column at x has a mass Am = (density * volume) = y(x) p t Ax. You add all the Am values to get the total mass M. The sum becomes an integral: M = pt y(2). For the center of mass, you add each column's x Am, and divide by M: pt M xy(x) dx Calculate xe. The only quantity you'll need is e = 5 m.
Transcribed Image Text:Do Chapter 12, Problem 5. This is an integration problem, to calculate the center of mass (center of gravity) for a continuous distribution. Specifically, we need the horizontal component of the center of mass. The height varies from h to zero according to this function: y(x) = h ( - 1). The constants h and e replace 1.00 m and 3.00 m. There is also a thickness t and a density p. You need two integrals, the total mass and the center of mass. Possibly surprisingly, you don't actually need the numbers t, h, and p. Ax y(x) The column at x has a mass Am = (density * volume) = y(x) p t Ax. You add all the Am values to get the total mass M. The sum becomes an integral: M = pt y(2). For the center of mass, you add each column's x Am, and divide by M: pt M xy(x) dx Calculate xe. The only quantity you'll need is e = 5 m.
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