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Physics / Physics with MasteringPhysics 4 / Chapter 9 / Problem 53P

Physics with MasteringPhysics | 4th Edition | ISBN: 9780321541635 | Authors: James S. Walker

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· · IP Three uniform meter sticks, each of mass m, are

Chapter 9, Problem 53P

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QUESTION:

Problem 53P

· · IP Three uniform meter sticks, each of mass m, are placed on the floor as follows: stick 1 lies along the y/axis from y = 0 to y = 1.0 m, stick 2 lies along the x axis from x = 0 to x = 1.0 m, stick 3 lies along the x axis from x = 1.0 m to x = 2.0 m. (a) Find the location of the center of mass of the meter sticks. (b) How would the location of the center of mass be affected if the mass of the meter sticks were doubled?

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QUESTION:

Problem 53P

· · IP Three uniform meter sticks, each of mass m, are placed on the floor as follows: stick 1 lies along the y/axis from y = 0 to y = 1.0 m, stick 2 lies along the x axis from x = 0 to x = 1.0 m, stick 3 lies along the x axis from x = 1.0 m to x = 2.0 m. (a) Find the location of the center of mass of the meter sticks. (b) How would the location of the center of mass be affected if the mass of the meter sticks were doubled?

ANSWER:

Step 1 of 2:-

Here we have to find the location of the center mass of the meter sticks system.

There are 3 meter sticks whose respective locations are,

$y=0\ to\ y=1\ m$ .

$x=0\ to\ x=1\ m$ .

$x=1\ m\ to\ 2\ m$ .

We know that,

${R}_{cm}=\frac{1}{M}\sum\limits_{i}^{\ }{m}_{i}{r}_{i}$

${X}_{cm}=\frac{1}{m+m+m}\times{}\left({m{x}_{1}+{m{x}_{2}+m{x}_{3}}_{\ }}\right)$

$=\frac{1}{3}\times{}\left({0+0.5+1.5}\right)$

$=0.66\ m$

Similarly,

${Y}_{cm}=\frac{1}{m+m+m}\times{}\left({m{y}_{1}+{m{y}_{2}+m{y}_{3}}_{\ }}\right)$

$=\frac{1}{3}\times{}\left({0.5+0+0}\right)$

$=0.16\ m$

So, the location of center of mass is $\left({0.66\ m,\ 0.16\ m}\right)$ .

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