Question

problem 7:
a meter stick of mass ( m = 100 g ) has a weight of ( m = 50 g ) attached at the far right end of the ruler. find the center of mass of the weight - meter stick system ( (x_{cm}=?) ). hint: where is the center of mass of a uniform meter stick?

Explanation:

Step1: Determine the position of the center of mass of the meter - stick and the weight

• A uniform meter - stick has its center of mass at \(x_{M}=50\ cm\) (mid - point of the meter - stick).
• The weight \(m = 50\ g\) is attached at the far - right end, so \(x_{m}=100\ cm\).
• The mass of the meter - stick \(M = 100\ g\).

Step2: Use the formula for the center of mass of a two - particle system

The formula for the center of mass \(X_{cm}=\frac{Mx_{M}+mx_{m}}{M + m}\).
Substitute \(M = 100\ g\), \(x_{M}=50\ cm\), \(m = 50\ g\), and \(x_{m}=100\ cm\) into the formula:

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Answer:

\(X_{cm}=\frac{200}{3}\ cm\approx66.7\ cm\)