A thin rod of length L and mass m has a linear density X(x) = Ax² where A is a constant and is the distance from the rod's left end. Locate the rod's center of mass (from x = = 0) in terms of its length by filling in the missing factor below.

College Physics
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Chapter1: Units, Trigonometry. And Vectors
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**Computation**

A thin rod of length \( L \) and mass \( m \) has a linear density \( \lambda(x) = Ax^2 \) where \( A \) is a constant and \( x \) is the distance from the rod's left end. Locate the rod’s center of mass (from \( x = 0 \)) in terms of its length by filling in the missing factor below.

[Hint: In terms of \( A \) and \( L \), the rod’s mass is \( M = \int_0^L \lambda \, dx \). Evaluate that integral and substitute into the denominator of the center of mass definition: 

\[ x_{cm} = \frac{\int_0^M x \, dm}{\int_0^M \, dm} = \frac{\int_0^L x \lambda(x) \, dx}{M} \]

The constant \( A \) will cancel and you’ll be left with an answer proportional to \( L \).]

\[ x_{cm} = \, \_\_\_\_\_ \, L \]

Record your numerical answer below, assuming three significant figures. Remember to include a “-" when necessary. 

[Blank space provided for answer]
Transcribed Image Text:**Computation** A thin rod of length \( L \) and mass \( m \) has a linear density \( \lambda(x) = Ax^2 \) where \( A \) is a constant and \( x \) is the distance from the rod's left end. Locate the rod’s center of mass (from \( x = 0 \)) in terms of its length by filling in the missing factor below. [Hint: In terms of \( A \) and \( L \), the rod’s mass is \( M = \int_0^L \lambda \, dx \). Evaluate that integral and substitute into the denominator of the center of mass definition: \[ x_{cm} = \frac{\int_0^M x \, dm}{\int_0^M \, dm} = \frac{\int_0^L x \lambda(x) \, dx}{M} \] The constant \( A \) will cancel and you’ll be left with an answer proportional to \( L \).] \[ x_{cm} = \, \_\_\_\_\_ \, L \] Record your numerical answer below, assuming three significant figures. Remember to include a “-" when necessary. [Blank space provided for answer]
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