Where is the Center of Gravity for a System of Attached Balls?

[lollypop]

hey all:
A ball with radius 0.081 m and mass 1.03kg is attached by a rod of length 4.5 m to a second ball with radius 0.1m and mass 1.94 kg . Suppose the rod is uniform with mass 1.53 kg .
Where is the system's center of gravity? Express answer as a distance measured from the center of the small ball.

I used the formula (m1d1+m2d2) / m1+m2, and it's wrong.i get confused since sometimes i see that they use a zero for m1d1, why is that?. and.. do i have to use the mass of the rod??

[?]

 

[Doc Al]

Originally posted by lollypop
I used the formula (m1d1+m2d2) / m1+m2, and it's wrong.i get confused since sometimes i see that they use a zero for m1d1, why is that?. and.. do i have to use the mass of the rod??

(1) Yes, you have to include the mass of the rod. Since the rod is uniform, its center of mass is at its middle.
(2) Since you are measuring position relative to the small ball, the position of the small ball is d1 = 0.

 

[expscv]

?? gravity between two balls? with such mass ?

and such distace? it seems not noticable


i think it is ask the gravity/ atrraction force by the rod to either the ball.

use f= ( G*m1*m2 )/r^2 r is the distance apart for two object

G= 6.67*10^-11 m1= mass 1 m2=mass 2

 

[Chen]

Originally posted by expscv
?? gravity between two balls? with such mass ?

The question asks for the center of gravity of the given system...

 

[expscv]

opps stupid me

 

[lollypop]

where exactly in the formula would i put the mass of the rod??

 

[Chen]

Originally posted by lollypop
where exactly in the formula would i put the mass of the rod??

Can't the formula be expanded to more than two masses?

$$d = \frac{m_1d_1 + m_2d_2 + m_3d_3}{m_1 + m_2 + m_3}$$

 

[lollypop]

i'm just not getting the right answer :(
i'm using zero as the firt distance , then i plug everything else and it's wrong, I've used the formula in every possible way.
this what i use :
(0 + m2d2+m3d3) / m1+m2+m3
am i supposed to use as d2 = r1+r2+rod? since the answer has to be measured form the center of the small ball??

 

[Chen]

Let $$l$$ be the length of the rod.

The position of the small ball's center of gravity (CG) relative to the small ball's center is zero, therefore $$d_1 = 0$$.

The position of the large ball's CG, which is located at its center, relative to the small ball's center is the length of the rod, plus the two radii of the balls, therefore $$d_2 = r_1 + r_2 + l$$.

The position of the rod's CG, which is located at its middle (since its mass is also evenly distributed), relative to the small ball's center is half of the rod's length plus the radius of the $$d_3 = \frac{l}{2} + r_1$$.

Your answer should, if my calculations are right, be that the center of gravity of the whole system is 2.81 meters from the small ball's center (and on the rod).

 

[lollypop]

thanks chen, my confusion was in the rod's center of gravity,
i didn't divide by 2, thanks again