Explanation Refer figure P5.39P and figure given below. Write the equation of curved boundary of shaded region. x 2 a 2 + y 2 b 2 = 1 Here x is x-coordinate of point on the curve, y is the y co-ordinate of the point on curve, a is the length of rectangle on P5.39, and b is the breadth of rectangle on P5.39. Rewrite the above equation in terms of y . y = b a a 2 − x 2 The shaded area in the above figure is a differential area element. Write the expression for the x-coordinate of center of mass of rectangular area element. x ¯ E L = x Here, x ¯ E L is the center of mass of differential area element d A . Write the expression for the y-coordinate of center of mass of d A . y ¯ E L = 1 2 ( y + b ) Here, y ¯ E L is the y-coordinate of center of mass of element d A . Rewrite the above expression by substituting b a a 2 − x 2 for y . y ¯ E L = 1 2 ( b a a 2 − x 2 + b ) = b 2 a ( a + a 2 − x 2 ) Write the expression for area of differential element. d A = ( b − y ) d x Here, d A is the area of rectangular area element and d x denotes a small change in x . Rewrite the above expression by substituting b a a 2 − x 2 for y . d A = ( b − b a a 2 − x 2 ) d x = b a ( a − a 2 − x 2 ) d x To find the area of shaded region in P5.39, integrate the right hand side of above equation from x = 0 to x = a . A = ∫ 0 a b a ( a − a 2 − x 2 ) d x (I) Here, A is the area of shaded region in figure P5.39. Assume the value a sin θ for x . x = a sin θ Differentiate the above relation with respect to x . 1 = a cos θ ( d θ d x ) Rewrite the above relation in terms of d x . d x = ( a cos θ ) d θ Find a 2 − x 2 by substituting a sin θ for x . a 2 − x 2 = a 2 − ( a sin θ ) 2 = a cos θ Now, change the integrating variable from x to θ . Then, the range of integration x = 0 to x = a changes to θ = 0 to θ = π 2 . Rewrite equation (I) by replacing limits x = 0 to x = a with θ = 0 to θ = π 2 , substituting ( a cos θ ) d θ for d x , and a cos θ for a 2 − x 2 . A = ∫ 0 π / 2 b a ( a − a cos θ ) ( a cos θ ) d θ = b a [ a 2 sin θ − a 2 ( θ 2 + sin 2 θ 4 ) ] 0 π / 2 = a b ( 1 − π 4 ) To calculate ∫ x ¯ E L d A , substitute b a ( a − a 2 − x 2 ) d x for d A and x for x ¯ E L and integrate it from x = 0 t0 x = a

12th Edition

ISBN: 9781260804638

Author: BEER

Publisher: MCG

See similar textbooks

Concept explainers

Center of Gravity and Centroid

The term center of gravity indicates the position or location of a point about which the whole weight of a body or system is supposed to be concentrated, which means the whole weight of the body or system works through this point. It is closely related t…

Videos

Question

Chapter 5.2, Problem 5.39P

To determine

The centrod of shaded area in Fig. P5.39.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 5.2, Problem 5.38P

Chapter 5.2, Problem 5.40P

Students have asked these similar questions

Locate the centroid of the plane area shown. G 1 = 72 mm 2= 120 mm Fig. P5.12

Locate the centroid of the plane area shown. r= 10 in. B a = 5 in. a = 5 in. Fig. P5.5

Locate the centroid of the plane area shown

Chapter 5 Solutions

VECTOR MECHANICS FOR ENGINEERS W/CON >B

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

The centrod of shaded area in Fig. P5.39.

Solution Summary: The author illustrates the centroid of shaded area in Fig. P5.39.

Concept explainers

Videos

The centrod of shaded area in Fig. P5.39.

Want to see the full answer?

Chapter 5 Solutions