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Center Of Mass Of The Uniform Rod - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • 7 Questions around this concept.

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A uniform rod of length 30cm is lying on a horizontal plane. If its opposite ends are moving with speed 1m/s and 2m/s respectively in the opposite directions. Then the instantaneous axis of rotation will be at a distance of xcm from the axis of rotation of its right end, moving with a speed of 2m/s. Then \frac{x}{l} will be

 

Concepts Covered - 1

Center of mass of the uniform rod

Suppose a rod of mass M and length L is lying along the x-axis with its one end at x = 0 and the other at x = L

                 

Mass per unit length of the rod = \mu = \frac{M}{L}

 

Take a small dx length of rod at a distance x from x=0

So mass of that dx element is =  dm = \mu.dx

Therefore, x-coordinate of COM of the rod will be 

x_{cm} = \frac{\int_{0}^{L}x.dm}{\int_{0}^{M}dm}

x_{cm} = \frac{\int_{0}^{L}x.\mu.dx}{\int_{0}^{M}dm} = \frac{\int_{0}^{L}x.\frac{M}{L}.dx}{\int_{0}^{M}dm}

x_{cm} = \frac{\frac{M}{L}\int_{0}^{L}xdx}{M} = \frac{1}{L}\int_{0}^{L}x.dx = \frac{L}{2}

So x coordinate of centre of mass of Uniform rod of length L 

 At a distance  \frac{L}{2}  from one of the ends  of the rod.   

Similarly  , y_{cm}= \frac{\int ydm}{\int dm}

And  y-coordinate is zero for all particles of rod

So,  y_{cm}= 0

Similarly,  z_{cm}= \frac{\int zdm}{\int dm}

And  z-coordinate is zero for all particles of rod

So,  z_{cm}= 0

So the coordinates of COM of the rod are (\frac{L}{2},0,0) 

Means it lies at the centre of the rod.

Position of centre of mass for uniform rectangular, square and circular plate
  1. Rectangular plate

  1. Square plate

  1. Circular plate

 

  1. For a 2-dimensional body with uniform negligible thickness formulae for finding the position of centre of mass can be rewritten  as

r_{cm}=\frac{m_1\vec{r}_1+m_2\vec{r}_2....}{m_1+m_2....}=\frac{ \rho A_1t\vec{r}_1+ \rho A_2t\vec{r}_2....}{ \rho A_1t+\rho A_2t....}=\frac{ A_1\vec{r}_1+ A_2\vec{r}_2....}{ A_1+ A_2....}

 

Where,  m = \rho.A.t

 

  1. Centre of mass when some mass is added in the body

{\vec{r}_{cm}}=\frac{m_{1}\vec{r_{1}}+{m_{2}}\vec{r}_{2}}{m_{1}+m_{2}}

Where \vec{r_{1}} are mass and position of the centre of mass for the whole body. & \vec{r_{2}} are mass and position of the centre of mass of added mass.

  1. Position of centre of mass when some mass is removed

                        {\vec{r}_{cm}}=\frac{m_{1}\vec{r_{1}}-{m_{2}}\vec{r}_{2}}{m_{1}-m_{2}}

 

Where m_1is value of whole mass and \vec{r_{1}}is position of centre of mass for whole mass.Similarly m_2 & \vec{r_{2}} are values for mass  which has been removed.

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Position of centre of mass for uniform rectangular, square and circular plate

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Position of centre of mass for uniform rectangular, square and circular plate

Physics Part II Textbook for Class XI

Page No. : 146

Line : 42

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