Center of Mass as an Average Point

Q. The position vector of center of mass of 'n' particles is a weighted average of the position vectors of the particles making up the system.

1. True
2. False

Q. A 15 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 4T radian/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm². The elongation of the wire when the mass is at the lowest point is (g = 10 m/s²)

Q. formula for angle between centripetal acceleration and resul†an t acceleratin

Q. 67. The x and y coordinates of the particle at any time are x=5t-2t and y=10t respectively, where x and y are in meters and t in seconds. The accerlation of the particles at t=2s is

Q. The position vector of three particles of mass $m_1=3\text{kg},m_2=4\text{kg}$ and $m_3=1\text{kg}$ are $\overrightarrow{r_1}=(2\hat{i}+\hat{j}+3\hat{k})\text{m},\overrightarrow{r_2}=(\hat{i}-3\hat{j}+2\hat{k})\text{m}$ and $\overrightarrow{r_3}=(3\hat{i}-2\hat{j}-\hat{k})\text{m}$ respectively. Find the position vector of centre of mass of the system of particles.

1. $\frac{1}{8}(13\hat{i}-12\hat{j}+15\hat{k})\text{m}$
2. $\frac{1}{8}(11\hat{i}-13\hat{j}+16\hat{k})\text{m}$
3. $\frac{1}{8}(16\hat{i}+11\hat{j}-13\hat{k})\text{m}$
4. $\frac{1}{8}(13\hat{i}-11\hat{j}+16\hat{k})\text{m}$

Q. the x and y coordinates of a particle are x=Asin(omegat) and y=Asin(omegat+pi/2).then motion of particle is

Q. Three particles $\text{A, B}$ & $\text{C}$ of masses $4m,2m$ & $3m$ are placed on the vertices of a equilateral triangle of length $2L$ as shown in the figure.

Find out the perpendicular distance of centre of mass from the side $\text{AB}$.

1. $\frac{L}{\sqrt{3}}$
2. $\sqrt{3}L$
3. $\frac{\sqrt{3}L}{2}$
4. $\frac{\sqrt{3}L}{4}$

Q. $n$ particles of the same mass are placed on the $x$ - axis. If the particles are placed at $x=1,2,3,..........n$ respectively, then find the location of COM.

1. $\frac{(n+1)}{2}$
2. $\frac{n(n+1)}{2}$
3. $\frac{n(n-1)}{2}$
4. None of the above

Q. The distance of the centre of mass of the T - shaped plate from O is nearly equal to

1. $7\text{m}$
2. $2.7\text{m}$
3. $4\text{m}$
4. $1\text{m}$

Q.

In HCl molecule, the separation between the nuclei of hydrogen and chlorine atoms is $1.27\stackrel{\circ }{A}$. If the mass of a chlorine atom is 35.5 times that of a hydrogen atom, the centre of mass of the HCl molecule is at a distance of

1. $\frac{35.5\times 1.27}{36.5}\stackrel{\circ }{A}$ from the hydrogen atom.

2. $\frac{35.5\times 1.27}{36.5}\stackrel{\circ }{A}$ from the chlorine atom.

3. $\frac{1.27}{36.5}\stackrel{\circ }{A}$ from the hydrogen atom.

4. $\frac{1.27}{36.5}\stackrel{\circ }{A}$ from the chlorine atom.

Q. Three particles of masses $1\text{kg}$ , $2\text{kg}$ and $3\text{kg}$ are placed at the vertices $A,B,C$ respectively of an equilateral triangle $ABC$ of edge $1\text{m}$ as shown in the figure. Find the position of the centre of mass of the system. (If $A$ is assumed to be the origin).

1. $(\frac{2}{3},\frac{\sqrt{3}}{6})$
2. $(\sqrt{\frac{3}{16}},\frac{2}{3})$
3. $(\frac{\sqrt{3}}{6},\frac{2}{3})$
4. $(\frac{2}{3},\sqrt{\frac{3}{16}})$

Q. In the $\text{HCl}$ molecule, the separation between the nuclei of the two atoms is about $1.5\stackrel{\circ }{A}(1\stackrel{\circ }{A}={10}^{-10}\text{m})$. The approximate location of the centre of mass from the hydrogen atom assuming the chlorine atom to be about 35.5 times as massive as hydrogen is

1. $1.45{\text{A}}^{\circ }$
2. $0.05{\text{A}}^{\circ }$
3. $0.72{\text{A}}^{\circ }$
4. $0.96{\text{A}}^{\circ }$

Q. The position of the centre of mass of a system consisting of two particles of masses $m_1$ and $m_2$, separated by distance $L$, from $m_2$ is

1. $\left(\frac{m_{2}L}{m_1+m_2}\right)$
2. $\left(\frac{m_1{m}_{2}L}{m_1^2+m_2^2}\right)$
3. $\left(\frac{m_1}{m_2}L\right)$
4. $\left(\frac{m_{1}L}{m_1+m_2}\right)$

Q.
Where is the Center of Mass of the 3 particle system shown in above figure?

1. $X_{cm}=1.15Y_{cm}=1.5$
2. $X_{cm}=1.07Y_{cm}=1.33$
3. $X_{cm}=1.5Y_{cm}=1.6$
4. None of these

Q. Three identical spheres each of mass $1\text{kg}$ are kept as shown in the figure. They are kept in such a way that they are touching each other with their respective centres on a straight line . If their centres are marked $P,Q,R$ respectively, then the distance of the centre of mass of the system from $P$ is:

1. $\frac{PQ+QR+PR}{3}$
2. $\frac{PQ+PR}{3}$
3. $\frac{PQ+PR}{2}$
4. $\frac{PQ+QR}{2}$

Q. Which of the following points is the likely position of the centre of mass of the system shown in the figure ?

1. A
2. B
3. C
4. D

Q. The centre of mass of a system of particles (arranged in a straight line) is at the origin. From this we conclude that

1. The number of particles on positive $x-$ axis is equal to the number of particles on negative $x-$ axis
2. The total mass of the particles on positive $x-$ axis is same as the total mass on negative $x-$ axis
3. The number of particles on $x-$ axis should be equal to the number of particles on $y-$ axis.
4. If there is a particle on the positive $x-$ axis, there must be at least one particle on the negative $x-$ axis.

Q. The average weight of 8 persons increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. The weight of the new person is .

1. 76 kg
2. 85 kg
3. 92 kg
4. 80.5 kg

Q. The position vector of four identical masses of mass $1\text{kg}$ each are $\overrightarrow{r_1}=(\hat{i}+2\hat{j}+7\hat{k}),\overrightarrow{r_2}=(3\hat{i}+5\hat{j}+\hat{k}),$ $\overrightarrow{r_3}=(6\hat{i}+2\hat{j}+3\hat{k})$ and $\overrightarrow{r_4}=(2\hat{i}-\hat{j}+5\hat{k})$. Find the position vector of their centre of mass.

1. $(2\hat{i}+3\hat{j}+4\hat{k})$
2. $(4\hat{i}+3\hat{j}+2\hat{k})$
3. $(3\hat{i}+4\hat{j}+2\hat{k})$
4. $(3\hat{i}+2\hat{j}+4\hat{k})$

Q. The mass per unit length of a non-uniform rod $OP$ of length $L$ varies as $m=k\frac{x}{L},$ where $k$ is a constant and $x$ is the distance of point on the rod from end $O$. The distance of the centre of mass of the rod from end $O$ is

1. $\frac{L}{3}$
2. $\frac{2L}{3}$
3. $\frac{L}{2}$
4. $\frac{2L}{\sqrt{3}}$

Q. Three particles of masses $2\text{kg}$, $3\text{kg}$ and $5\text{kg}$ are placed at the three vertices $A,B,C$ of a right angled triangle respectively. Co-ordinates of the particles are $A(0,3),B(0,0)$ and $C(3,0)$. Find the position of centre of mass of the particles.

1. $(1,1)$
2. $(2,1)$
3. $(1.5,0.6)$
4. $(1.5,1.6)$

Q. Derivation of center of mass in quadrant of circle

Q. The position of the centre of mass of a system consisting of two particles of masses $m_1$ and $m_2$, separated by distance $L$, from $m_2$ is

1. $\left(\frac{m_{2}L}{m_1+m_2}\right)$
2. $\left(\frac{m_1{m}_{2}L}{m_1^2+m_2^2}\right)$
3. $\left(\frac{m_1}{m_2}L\right)$
4. $\left(\frac{m_{1}L}{m_1+m_2}\right)$

Q.

In HCl molecule, the separation between the nuclei of hydrogen and chlorine atoms is $1.27\stackrel{\circ }{A}$. If the mass of a chlorine atom is 35.5 times that of a hydrogen atom, the centre of mass of the HCl molecule is at a distance of

1. $\frac{35.5\times 1.27}{36.5}\stackrel{\circ }{A}$ from the hydrogen atom.

2. $\frac{35.5\times 1.27}{36.5}\stackrel{\circ }{A}$ from the chlorine atom.

3. $\frac{1.27}{36.5}\stackrel{\circ }{A}$ from the hydrogen atom.

4. $\frac{1.27}{36.5}\stackrel{\circ }{A}$ from the chlorine atom.

Q. The position vector of four particles of masses $m_1=2\text{kg},m_2=4\text{kg},m_3=6\text{kg},m_4=8\text{kg}$ are $\overrightarrow{r_1}=2\hat{i}+3\hat{j}+0\hat{k},\overrightarrow{r_2}=0\hat{i}+2\hat{j}+3\hat{k},\overrightarrow{r_3}=2\hat{i}+0\hat{j}+3\hat{k};\overrightarrow{r_4}=2\hat{i}+3\hat{j}+3\hat{k}$. The position vector of their centre of mass is given by

1. $\frac{16\hat{i}+19\hat{j}+27\hat{k}}{11}$
2. $1.6\hat{i}+1.9\hat{j}+2.7\hat{k}$
3. $8\hat{i}+9\hat{j}+13\hat{k}$
4. $\frac{32\hat{i}+38\hat{j}+54\hat{k}}{11}$

Q. Four point masses $2\text{kg},3\text{kg},5\text{kg}$ and $7\text{kg}$ respectively are placed at the four corners $A,B,C$ and $D$ of a square of side $4\text{m}$. The position of $COM$ will be

1. $(\frac{48}{17},\frac{32}{17})\text{m}$
2. $(\frac{32}{17},\frac{48}{17})\text{m}$
3. $(2,4)$
4. $(2,2)$

Q. A uniform metal disc of radius R is taken and out of it a disc of diameter R is cut off from the end. The centre of mass of the remaining part will be

1. $\frac{R}{4}$ from the centre
2. $\frac{R}{3}$ from the centre
3. $\frac{R}{5}$ from the centre
4. $\frac{R}{6}$ from the centre

Q. A rigid body consists of a $3\text{kg}$ mass connected to a $4\text{kg}$ mass by a massless rod. The $3\text{kg}$ mass is located at $\overrightarrow{r_1}=(3\hat{i}+2\hat{j})\text{m}$ and the $4\text{kg}$ mass at $\overrightarrow{r_2}=(8\hat{i}+7\hat{j})\text{m}$. Find the length of the rod and the co-ordinates of the centre of mass

1. $5\text{m},(\frac{41}{7},\frac{34}{7})\text{m}$
2. $5\sqrt{2}\text{m},(\frac{41}{7},\frac{34}{7})\text{m}$
3. $5\text{m}$, $(6.5,0)\text{m}$
4. $5\sqrt{2}\text{m},(5,0)\text{m}$

Q. Three particles of masses $2\text{kg}$, $3\text{kg}$ and $5\text{kg}$ are placed at the three vertices $A,B,C$ of a right angled triangle respectively. Co-ordinates of the particles are $A(0,3),B(0,0)$ and $C(3,0)$. Find the position of centre of mass of the particles.

1. $(1,1)$
2. $(2,1)$
3. $(1.5,0.6)$
4. $(1.5,1.6)$

Q. A rigid body consists of a $3\text{kg}$ mass connected to a $4\text{kg}$ mass by a massless rod. The $3\text{kg}$ mass is located at $\overrightarrow{r_1}=(3\hat{i}+2\hat{j})\text{m}$ and the $4\text{kg}$ mass at $\overrightarrow{r_2}=(8\hat{i}+7\hat{j})\text{m}$. Find the length of the rod and the co-ordinates of the centre of mass

1. $5\text{m},(\frac{41}{7},\frac{34}{7})\text{m}$
2. $5\sqrt{2}\text{m},(\frac{41}{7},\frac{34}{7})\text{m}$
3. $5\text{m}$, $(6.5,0)\text{m}$
4. $5\sqrt{2}\text{m},(5,0)\text{m}$