Question

The linear mass density of a rod of length 5 m varies with distance from its end fixed at the origin as represented in the graph shown below:

Then find the position of centre of mass of the rod w.r.t origin.

Then find the position of centre of mass of the rod w.r.t origin.

- 2516 m
- 59 m
- 259 m
- 516 m

Solution

The correct option is **C** 259 m

From the graph of linear mass density of rod, writing it in y=mx+c format:

λ=(5+x) kg/m

Let us consider a small element of length dx and mass dm, which is at a distance x from the origin.

Let xCM be the position of COM of rod, so by applying basic formula and integrating with proper limits:

xCM=∫x dm∫dm...(i)

λ=dmdx=5+x

⇒dm=(5+x)dx....(ii)

Substituting dm in Eq (i) and setting the values of limits as x=0 to x=5:

xCM=∫50x(5+x)dx∫50(5+x)dx=∫50(5x+x2)dx∫50(5+x)dx

=[5x22+x33]50[5x+x22]50

=[5(5)22+(5)33][5(5)+(5)22]

⇒xCM=259 m

From the graph of linear mass density of rod, writing it in y=mx+c format:

λ=(5+x) kg/m

Let us consider a small element of length dx and mass dm, which is at a distance x from the origin.

Let xCM be the position of COM of rod, so by applying basic formula and integrating with proper limits:

xCM=∫x dm∫dm...(i)

λ=dmdx=5+x

⇒dm=(5+x)dx....(ii)

Substituting dm in Eq (i) and setting the values of limits as x=0 to x=5:

xCM=∫50x(5+x)dx∫50(5+x)dx=∫50(5x+x2)dx∫50(5+x)dx

=[5x22+x33]50[5x+x22]50

=[5(5)22+(5)33][5(5)+(5)22]

⇒xCM=259 m

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