Question

A meter stick is pivoted at its 50 cm mark but does not balance because of non-uniformities in its material that cause its center of gravity to be displaced from its geometrical center. However, when weights of 150 g and 300 g are placed at the 10 cm and 75 cm marks, respectively, balance is obtained. The weights are then interchanged and balance is again obtained by shifting the pivot point to the 43 cm mark. Find the mass of the meter stick and the location of its center of gravity.

Answer 1

Answer:

The mass of the stick is:

$M=0,9428kg$

The center of mass is at the 48,41cm mark

Explanation:

To be balanced in any case the resultant torque of the system must be zero.

For the first situation:

We have a 0,15kg mass al -0,4m from the pivot point, and a 0,3kg mass at +0,25m of the pivot point. On the other hand we know that the center of gravity of the stick is not at the 50cm mark point, so it will be at a X distance with a mass M, the torque sum:

$0,15kg*(-0,4m)+0,3kg*0,25m+X*M=0$

So:

(1)$X*M=-0,015kgm$

For the second situation:

Now the weights are interchanged, and as the pivot point has change, the distances from the pivot point are different:

We have a 0,3kg mass al -0,33m from the pivot point, and a 0,15kg mass at +0,32m of the pivot point. As the pivot has moved we have to move our reference for X: The center of gravity of the stick is at a X+0,07m distance with a mass M, the torque sum:

$0,3kg*(-0,33m)+0,15kg*0,32m+(X+0,07m)*M=0$

$(X+0,07m)*M=0,051kgm$

$X*M+0,07m*M=0,051kgm$

replacing X*M from (1):

$-0,015kgm+0,07m*M=0,051kgm$

The mass of the stick is:

$M=(0,051kgm+0,015kgm)/0,07m=0,9428kg$

and replacing this in (1):

$X*0,9428kg=-0,015kgm$

$X=-0,015kgm/0,9428kg=-0,0159m=-1,59cm$

As we took the 50cm mark as the reference, the center of mass is at the 48,41cm  mark