Lazydog\'s friend has a power T medal as shown above. The T is of uniform thickn

Question

Lazydog's friend has a power T medal as shown above. The T is of uniform thickness and density. All dimensions are given in inches. The center of the hole is offset 1.85 from the top and the left, and has a diameter of 3.30.

A.   Neglecting the hole, what is the horizontal distance from the left side of the T to its CG?
(include units with answer)  

B.   Neglecting the hole, what is the vertical distance from the top of the T to its CG?
(use downwards as positive)
(include units with answer)  

C.   Now include the hole - what is the surface area of the T?
(include units with answer)  

D.   What is the horizontal distance from the left side of the T to its CG?
(include units with answer)  

E.   What is the vertical distance from the top of the T to its CG?
(use downwards as positive)
(include units with answer)

Explanation / Answer

A) about middle axis T shape is identical ,

so x coordinate of CG will be on middle axis,

distance of CG from left = (11.1/2) = 5.55

B)
using

Centre of mass Ycm = ( m1y1 + m2y2 + m3y3 + ..... ) / (m1 + m2 + m3 + .... )

suppose area density is & then mass = area x density

Ycm = ( (8.4 x 2 x & x (3.7+5.4+1) + ( 3.9x 5.4 x & x (3.7 + (5.4/2))) + ( 11.1 x 3.7 x & x (3.7/2))) / ( (8.4 x 2 x &) + ( 3.9x 5.4 x &) + ( 11.1 x 3.7 x&) )

Ycm = 4.82 from top.

C) surafce area of hole = pi (3.30/2)^2 = 8.55

d) now, new Xcm = (mass x xcm) - (8.55 x & x 1.85) / (mass - 8.55&)    78.93

      = ( 5.55 x 78.93&) - (15.82&) / (78.93 - 8.55)&

     = 6

e) now, new ycm = (mass x Ycm) - (8.55 x & x 1.85) / (mass - 8.55&)    78.93

      = ( 4.82 x 78.93&) - (15.82&) / (78.93 - 8.55)&

     = 5.18

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