Finding the Mass of a Suspended Necklace Using the Center of Mass Method

[vworange]

Been trying this one using:
x1m1 = x2m2

I guess I just maybe am not using the right mass or length for center of mass..

Here's the question:

A 0.19 kg meter stick balances at its center. If a necklace is suspended from one end of the stick, the balance point moves 16.3 cm toward that end. What is the mass of the necklace?

 

[physicsCU]

You need to use calculus, not physics. Look up center of mass of a rod on google.

I don't remember the formulas, that was like two months ago.

 

[vincentchan]

don't mislead him, no calculas is needed in this problem, the center of mass of a rod is at its mid-point...
here is some hint:
draw the graph first, and find the torque for LHS and RHS... the torque of the necklace is straight forward, but the torque for the rod is a bit tricky...

 

[Andrew Mason]

Been trying this one using:
x1m1 = x2m2

I guess I just maybe am not using the right mass or length for center of mass..

Here's the question:

A 0.19 kg meter stick balances at its center. If a necklace is suspended from one end of the stick, the balance point moves 16.3 cm toward that end. What is the mass of the necklace?

The stick has a mass/length of .19 kg/m or 1.9 g/cm.=\rho The torque when the necklace is added and the balance point is moved is 0:

M_nd_1 + \rho d_1^2/2 - \rho d_2^2/2 = 0

d_1 = 33.7 and d_2 = 66.3 cm

(you can think of the torque from one side of the stick as a point mass sitting half way between the end and the balance point on a massless stick).

AM

 

[Touchkin]

There's no need to divide the rod into 2 pieces and to calculate the torque of each of them.

This problem is as simple as 1,2,3. :zzz:

Look:

Write the equation of the torques around the new ballance point:

15*0.2=35*m
m~0.09kg

That's it!
Enjoy physics!