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The center of mass is located at position i EX.5.4. Locate the center of mass of a uniform rod of mass M and length

Answer» <html><body><p></p>Solution :Consider a uniform <a href="https://interviewquestions.tuteehub.com/tag/rod-1190628" style="font-weight:bold;" target="_blank" title="Click to know more about ROD">ROD</a> of mass M and length / whose one <a href="https://interviewquestions.tuteehub.com/tag/end-971042" style="font-weight:bold;" target="_blank" title="Click to know more about END">END</a> coincides with the origin as shown in Figure. The rod is kept along the x <a href="https://interviewquestions.tuteehub.com/tag/axis-889931" style="font-weight:bold;" target="_blank" title="Click to know more about AXIS">AXIS</a>. To find the <a href="https://interviewquestions.tuteehub.com/tag/center-11455" style="font-weight:bold;" target="_blank" title="Click to know more about CENTER">CENTER</a> of mass of this rod, we choose an infinitesimally small mass dm of elemental length dr at a <a href="https://interviewquestions.tuteehub.com/tag/distance-116" style="font-weight:bold;" target="_blank" title="Click to know more about DISTANCE">DISTANCE</a> from the origin <br/> `lambda`is the linear mass density (ie, mass per unit length) of the rod. `lambda=(M)/(l)` <br/> The mass of small element (dm) is `dm, (M)/(l)dx` <br/> Now, we can write the center of mass equation for this mass distribution as, <br/> `X_(CM)=(int x dxm)/( int dm)` <br/> `X_(CM)=( overset(1) underset(0) intx((M)/(l)dx))/(M)=(1)/(l)overset(1) underset(0) intxdx=(1)/(l)[(x^(2))/(2)]_(0)^(l)=(1)/(l)=((t^(2))/(2))` <br/> `X_(CM)=(l)/(2)` <br/> As the `(l)/(2)` position is the geometric center of the rod, it is concluded that the center of mass <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_PHY_XI_V01_C05_SLV_004_S01.png" width="80%"/></body></html>


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