NEET PG 2024 Cut off Reduce by NBEMS - Qualifying Percentile & Branch-wise Expected Cutoff

Center Of Mass Of The Uniform Rod MCQ - Practice Questions with Answers

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

Quick Facts

9 Questions around this concept.

Solve by difficulty

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

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{1}{2}$

$\frac{3}{4}$

View Solution

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

Rectangular plate

Square plate

Circular plate

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, $m = \rho.A.t$

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.

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 $m_1$ is 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.