Applied Maths - Mechanics 8

Imagine the hole is on the left hand side of O in a diagram.
The disc is then attached to the circumference to the right hand side of O.
Let the new centre of mass be at x, with right hand side as the positive direction..
Take mass moment about the centre O along the axis of symmetry, using the concept of negative mass for the hole, and assume a density of p per unit area,
-πr2p(-R/2) + πR2p(0) + πr2p(R+r) = πR2px
[Note: 0 in the expression is due to O is the centre of mass of the original disc]
Eliminate π and p =>
r2(R/2) + r2(R+r) = R2x
x = (r2/R2)(R/2 + R + r)
= r2(3R + 2r)/(2R2)

Let p be the density of the disc.

For the circular ring,

Mass of the disc with radius R = p πR2

Mass of the disc with radius r = p πr2

Let (x , 0) be the centre of mass of the circular ring.

x = [(p πR2)(0) - (pπr2)(R/2)] / (pπR2- pπr2) = -Rr2 / 2(R2 - r2)


Now, consider the lamina

Let (a , 0) be the centre of mass of the lamina.

a = {(pπR2 - pπr2)[-Rr2 / 2(R2 - r2)]+ (pπr2)(R + r/2)} / pπR2

= (-Rr2/2 + Rr2 + r3/2) / R2

= r2(R + r) / R2

So, the required distance is r2(R + r) / R2

2009-07-27 16:14:08 補充：
Correction:

Consider the lamina, a = {(pπR^2 - pπr^2)[Rr^2 / 2(R^2 - r^2)]+ (pπr^2)(R + r/2)} / pπR^2

= (Rr^2/2 + Rr^2 + r^3/2) / R^2

= r^2(3R + 2r) / 2R^2