Informative line

### Velocity And Acceleration Of Center Of Mass

Find velocity of center of mass for 1 d motion & 2 d motion, acceleration of center of mass for Atwood machine 1 d & 2 d. Practice equation of momentum of centre of mass (COM) of a system.

# Velocity of Center of Mass for 1 D Motion

## Center of mass for n–particle system

• Position of center of mass for n–particle system is given by

$$\vec{r}_{cm} = \dfrac{m_1 \vec{r_1} + m_2\vec{r_2} + m_3\vec{r_3} + ......+ m_n \vec{r_n}}{m_1 +m_2 +m_3......+ m_n}$$

Here, $$m_1,m_2.....m_n$$ are masses of particles and $$\vec{r_1}, \;\;\vec{r_2},.....\vec{r_n}$$ are their respective position vectors.

or,

$$\vec{r_{cm }} = \dfrac{m_1\vec{r_1} + m_2\vec{r_2} + ....+ m_n\vec{r_n}}{\sum\limits_{i = 1}^{ n} \; m_i}\;\;\;\;\;\;\;\;\;— (1)$$

## Velocity of center of mass

• From equation (1)

$$\sum \; m_i (\vec{r_{cm}}) = m_1 \vec{r_1} + m_2 \vec{r_2} + ......+m_n \vec{r_n}$$

Differentiating the above equation on both sides

$$\sum \; mi (\vec{v}_{cm}) = m_1 \vec {v}_1 + m_2 \vec {v}_2 + .....m_n \vec {r}_n\;\;\;\;\;— (2)$$

Thus,   $$\vec{v}_{cm} = \dfrac{m_1 \vec {v}_1 + m_2 \vec {v}_2 + .... + m_n \vec {r}_n}{\sum\limits_{i= 1}^{n} \; m_i}\;\;\;\;\;— (3)$$

## Acceleration of Center of Mass

• Again differentiating equation (2)

$$\sum m_i (\vec{a}_{cm}) = m_1\vec{a_1} + m_2\vec{a_2} + ..... +m_n\vec{a_n}$$

or,

$$\vec{a_{cm}} = \dfrac{m_1\vec{a_1} + m_2\vec{a_2} + .... m_n\vec{a_n}}{\sum\limits_{i = 1} ^{ n} \; m_i}\;\;\;\;\;\;\;\;—(4)$$

#### Two bodies of masses 4 kg and 2 kg are moving with velocities 4 m/s and 8 m/s respectively on a smooth surface and in same direction as shown in figure. Find the velocity of center of mass.

A $$\dfrac{ 14}{3 } \hat{i} \; m/s$$

B $$\dfrac{ 18}{ 3} \hat{i} \; m/s$$

C $$\dfrac{ 16}{3 } \hat{i} \; m/s$$

D $$\dfrac{19 }{3 } \hat{i} \; m/s$$

×

Write $$\vec{v_1}\; \text and \;\vec{v_2}$$ , vectorially

$$\vec{v_1} = 4 \,\hat{i}\;m/s \;\;\;\text and\;\;\;\vec{v_2} = 8 \,\hat{i} \;m/s$$

Use equation of velocity of center of mass

$$\vec{v}_{cm} = \dfrac{m_1 \vec {v_1} + m_2 \vec{v_2}}{m_1 +m_2}$$

$$\vec{v}_{cm} = \dfrac{4 × 4\,\hat{i} + 2 × 8\,\hat{i} }{6} = \dfrac{32\,\hat{i} }{6}$$

$$\vec{v}_{cm} = \dfrac{16}{3}\,\hat{i} \, m/s$$

### Two bodies of masses 4 kg and 2 kg are moving with velocities 4 m/s and 8 m/s respectively on a smooth surface and in same direction as shown in figure. Find the velocity of center of mass.

A

$$\dfrac{ 14}{3 } \hat{i} \; m/s$$

.

B

$$\dfrac{ 18}{ 3} \hat{i} \; m/s$$

C

$$\dfrac{ 16}{3 } \hat{i} \; m/s$$

D

$$\dfrac{19 }{3 } \hat{i} \; m/s$$

Option C is Correct

# Velocity of Center of Mass for 2 D Motion

## Center of mass for n–particle system

• Position of center of mass for n–particle system is given by

$$\vec{r}_{cm} = \dfrac{m_1 \vec{r_1} + m_2\vec{r_2} + m_3\vec{r_3} + ......+ m_n \vec{r_n}}{m_1 +m_2 +m_3......+ m_n}$$

Here, $$m_1,m_2.....m_n$$ are masses of particles and $$\vec{r_1}, \;\;\vec{r_2},.....\vec{r_n}$$ are their respective position vectors.

or,

$$\vec{r_{cm }} = \dfrac{m_1\vec{r_1} + m_2\vec{r_2} + ....+ m_n\vec{r_n}}{\sum\limits_{i = 1}^{n} \; m_i}\;\;\;\;\;\;\;\;\;— (1)$$

## Velocity of center of mass

• From equation (1)

$$\sum \; m_i (\vec{r_{cm}}) = m_1 \vec{r_1} + m_2 \vec{r_2} + ......+m_n \vec{r_n}$$

Differentiating the above equation on both sides

$$\sum \; mi (\vec{v}_{cm}) = m_1 \vec {v}_1 + m_2 \vec {v}_2 + .....m_n \vec {r}_n\;\;\;\;\;— (2)$$

Thus,   $$\vec{v}_{cm} = \dfrac{m_1 \vec {v}_1 + m_2 \vec {v}_2 + .... + m_n \vec {r}_n}{\sum\limits_{i= 1}^{n} \; m_i}\;\;\;\;\;— (3)$$

• In 2–D motion of particles of a system, the velocity of center of mass is calculated for $$x$$ – axis and $$y$$ – axis separately, using $$x$$ – components and $$y$$ – components of corresponding velocities of particles.
• If $$v_{cm(x)}$$ and $$v_{cm(y)}$$ be the $$x$$ and $$y$$ – components of center of mass for n–particle system,then velocity of center of mass in $$x$$ – direction $$(\vec{v}_{cm(x)})$$ is given by

$$\vec{v}_{cm(x)} = \dfrac{m_1 \vec{v}_{x_1}+m_2 \vec{v}_{x_2} +.....+m_n \vec{v}_{x_n}}{\sum\; m}$$

• Velocity of center of mass in $$y$$– direction $$(\vec{v}_{cm(y)})$$ is given by

$$\vec{v}_{cm(y)} = \dfrac{m_1 \vec{v}_{y_1 } + m_2 \vec{v}_{y_2}+ .... + m_n \vec{v}_{y_n}}{\sum \; m}$$

or,

we can directly apply the formula,

$$\vec{v}_{cm} = \vec{v}_{cm(x)} + \vec{v}_{cm(y)}$$

$$\vec{v}_{cm} = \dfrac{m_1(\vec{v}_{x_1} + \vec{v}_{y_1}) + m_2(\vec{v}_{x_2} + \vec{v}_{y_2})+....+m_n (\vec{v}_{x_n} +\vec{v}_{y_n} )}{\sum \; m}$$

$$\vec{v}_{cm} = \dfrac{(m_1 \;v_{x_1} + m_2 \; v_{x_2}+ .... + m_n \; v_{x_n})\,\hat i}{\sum \; m} + \dfrac{(m_1 \;v_{y_1} + m_2 \; v_{y_2}+ .... + m_n \; v_{y_n})\,\hat j}{\sum \; m}$$

#### Velocities of two particles of mass $$m_1 = 1 \; kg$$ and $$m_2 = 4 \; kg$$ are given as $$\vec{v}_1 = (2 \hat i + 3 \hat j)\;m/sec\;\; and \;\;\vec{v}_2 = (3\hat i + 4 \hat j ) \,m/sec$$ respectively. The velocity of center of mass of the system will be

A $$(1.4 \;\hat i + 1.9 \;\hat j )\; m/sec$$

B $$(2.8 \;\hat i + 3.8 \;\hat j)\; m/sec$$

C $$(-1.4\;\hat i + 1. 9\;\hat j)\; m/sec$$

D $$( - 2.8 \;\hat i + 3. 8 \;\hat j) \; m/sec$$

×

Write velocity of each particle vectorially

$$\vec{v }_1 = (2 \;\hat i + 3 \;\hat j )\; m/sec$$

$$\vec{v}_2 = (3\;\hat i +4\;\hat j )\; m/sec$$

also, $$m = 1 \; kg$$ and $$m_2 = 4\; kg$$

Write $$x$$ and $$y$$ components of velocities for each particle, separately.

$$\vec{v}_{1x} = 2 \;\hat i \; m/s, \;\;\;\;\;\vec{v}_{1y} = 3 \;\hat j \; m/s$$

$$\vec{v}_{2x} = 3 \;\hat i \; m/s , \;\;\;\; \vec{v}_{2y} = 4\;\hat i \; m /s$$

Calculate velocity of center of mass in $$x$$ and $$y$$ direction separately using formulae

$$v_{cm(x)} = \dfrac{m_1 v_{1x} + m_2 v_{2 x} + ..... + m_n v_{nx}}{\sum\; m}$$

$$v_{cm(y)} = \dfrac{m_1v_{1y} + m_2v_{2y} + ....+ m_n v_{ny}}{\sum \; m}$$

Substituting values

$$v_{cm(x)} = \dfrac{1 (2) + 4 (3)}{1+4} = \dfrac{14}{5} = 2.8 \; m/s$$

$$v_{cm(y)} = \dfrac{1 (3)+ 4(4)}{1 +4} = \dfrac{19}{5} = 3.8\; m/s$$

Hence, velocity of center of mass $$\vec{v}_{cm} = v_{cm(x)}\;\hat i + v_{cm(y) }\;\hat j$$

$$\vec{v}_{cm} =( 2. 8 \;\hat i + 3. 8 \;\hat j) \; m/s$$

### Velocities of two particles of mass $$m_1 = 1 \; kg$$ and $$m_2 = 4 \; kg$$ are given as $$\vec{v}_1 = (2 \hat i + 3 \hat j)\;m/sec\;\; and \;\;\vec{v}_2 = (3\hat i + 4 \hat j ) \,m/sec$$ respectively. The velocity of center of mass of the system will be

A

$$(1.4 \;\hat i + 1.9 \;\hat j )\; m/sec$$

.

B

$$(2.8 \;\hat i + 3.8 \;\hat j)\; m/sec$$

C

$$(-1.4\;\hat i + 1. 9\;\hat j)\; m/sec$$

D

$$( - 2.8 \;\hat i + 3. 8 \;\hat j) \; m/sec$$

Option B is Correct

# Acceleration of Center of Mass for Atwood Machine(2 D)

• ## Center of mass for n–particle system

Position of center of mass for n–particle system is given by

$$\vec{r}_{cm} = \dfrac{m_1 \vec{r_1} + m_2\vec{r_2} + m_3\vec{r_3} + ......+ m_n \vec{r_n}}{m_1 +m_2 +m_3......+ m_n}$$

Here, $$m_1,m_2.....m_n$$ are masses of particles and $$\vec{r_1}, \;\;\vec{r_2},.....\vec{r_n}$$ are their respective position vectors.

or,

$$\vec{r_{cm }} = \dfrac{m_1\vec{r_1} + m_2\vec{r_2} + ....+ m_n\vec{r_n}}{\sum\limits_{i = 1}^{ n} \; m_i}\;\;\;\;\;\;\;\;\;— (1)$$

• ## Velocity of center of mass

From equation (1)

$$\sum \; m_i (\vec{r_{cm}}) = m_1 \vec{r_1} + m_2 \vec{r_2} + ......+m_n \vec{r_n}$$

Differentiating the above equation on both sides

$$\sum \; mi (\vec{v}_{cm}) = m_1 \vec {v}_1 + m_2 \vec {v}_2 + .....m_n \vec {r}_n\;\;\;\;\;— (2)$$

Thus,   $$\vec{v}_{cm} = \dfrac{m_1 \vec {v}_1 + m_2 \vec {v}_2 + .... + m_n \vec {r}_n}{\sum\limits_{i= 1}^{n} \; m_i}\;\;\;\;\;— (3)$$?

• ## Acceleration of Center of Mass

Again differentiating equation (2)

$$\sum m_i (\vec{a}_{cm}) = m_1\vec{a_1} + m_2\vec{a_2} + ..... +m_n\vec{a_n}$$

or,

$$\vec{a_{cm}} = \dfrac{m_1\vec{a_1} + m_2\vec{a_2} + .... m_n\vec{a_n}}{\sum\limits_{i = 1} ^{ n} \; m_i}\;\;\;\;\;\;\;\;—(4)$$

#### For the arrangement as shown, calculate the acceleration of center of mass of two blocks.  [ Take g = 10 m/s2 and assume surfaces to be frictionless ]

A $$(2.4\;\hat{i} - 3.6 \; \hat{j}) \; m/s^2$$

B $$(3.4\; \hat{i} - 6.3 \; \hat{j}) \; m/s^2$$

C $$(5.1 \;\hat{i} - 3.6 \; \hat{j}) \; m/s^2$$

D $$(3.5 \;\hat{i}-2.6\; \hat{j} )\; m/s^2$$

×

Find acceleration/velocity of individual masses in system.

Here, magnitude of acceleration of both blocks will be same.

$$a = \dfrac{\text{Net driving force }}{\text {Total mass}} = \dfrac{3g}{5} = \dfrac{3 × 10}{5} = 6 \; m/s^2$$

List the given velocities/accelerations vectorially, for each particle.

$$m_1 = 2 \; kg , \;\;\;\; \vec {a}_1 = 6 \hat{i} \; m/s^2$$

$$m_2 = 3 \; kg , \;\;\;\;\vec {a}_2 = - 6\; \hat{j}\;m/s^2$$

Write x and y–components of accelerations for each particle separately.

For particle 1, $$m_1 = 2 \; kg,\;\;\;\;\; a_{1x} = 6 \; m/s^2,\;\;\;a_{1y} = 0 \; m/sec^2$$

For particle 2, $$m_2 = 3 \; kg,\;\;\;\;\; a_{2x} = 0 \; m/s^2,\;\;\;a_{2y} = -6 \; m/sec^2$$

Calculate the acceleration of center of mass using formulae

$$a_{cm (x)} = \dfrac{m_1a_{1x} + m_2 a_{2 x} + ......+ m_na_{n x}}{\sum\; m}$$

$$a_{cm}(y) = \dfrac{m_1a_{1y} + m_2 a_{2y} + .....+ m_n a_{n y}}{\sum\; m}$$

Substituting values

$$a_{cm(x)} = \dfrac{2(6)+ 3 (0)}{2 +3 } = \dfrac{12}{5} = 2.4 \; m/s^2$$

$$a_{cm(y)} = \dfrac{2(0)+ 3 (-6) }{2+3} = \dfrac{-18}{5} = - 3. 6 \; m/s^2$$

$$\vec{a}_{cm} =( 2. 4\; \hat {i}- 3.6 \; \hat {j})\;m/s^2$$

### For the arrangement as shown, calculate the acceleration of center of mass of two blocks.  [ Take g = 10 m/s2 and assume surfaces to be frictionless ]

A

$$(2.4\;\hat{i} - 3.6 \; \hat{j}) \; m/s^2$$

.

B

$$(3.4\; \hat{i} - 6.3 \; \hat{j}) \; m/s^2$$

C

$$(5.1 \;\hat{i} - 3.6 \; \hat{j}) \; m/s^2$$

D

$$(3.5 \;\hat{i}-2.6\; \hat{j} )\; m/s^2$$

Option A is Correct

# Calculation of Momentum of Centre of Mass(COM) of Two Particles in 2 D

• ## Momentum of System of Particles in a System

Consider a n–particle system, having masses $$m_1 , m_2,........m_n$$ with velocities $$\vec{v}_1, \vec{v}_2, .....\vec{v}_n$$ respectively. Momentum of an individual particle will be given as

For particle 1 : $$\vec{p}_1 = m_1 \, \vec{v}_1$$

For particle 2 : $$\vec{p}_2 = m_2 \, \vec{v}_2$$

For nth particle : $$\vec{p}_n = m_n \,\vec{v}_n$$

$$m_1 + m_ 2 + .......m_n = \sum m = M$$

• ## Total Momentum of all Particles

$$\vec{p} = \vec{p}_1 + \vec{p}_2 +......+ \vec{p}_n$$ (Vector sum of individual momentum)

$$\vec{p} = m_1 \vec {v}_1 + m_2 \vec{v}_2 + ... ..+ m_n \vec{v}_n$$

• ## Velocity of Center of Mass $$(\vec{v}_{cm})$$

Velocity of COM is given by

$$\vec{v}_{cm} = \dfrac{m_1 \vec{v}_1 + m_2 \vec{v }_2 + .... . + m_n \vec{v}_n}{\sum \; m}$$

$$(\sum m) \vec{v}_{cm} = m_1 \vec {v}_1 + m_2 \vec{v}_2 + ..... .+ m_n \vec {v}_n$$

$$M\;\vec{v}_{cm} = \vec {p}_1 + \vec{p}_2 + ...... + \vec {p}_n$$

• Thus, total momentum of system is the vector sum of all the momentum of individual particles which is equal to the momentum of center of mass.

Hence, total momentum of system depends only upon the velocity of COM.

• To understand the concept better consider an example of a fan. When the fan is rotated, each particle on fan has momentum and it is very difficult to calculate total momentum of the system.
• But as we know velocity of COM is zero, so total momentum of fan will be zero, as $$v_{cm} = 0$$.

Note : - Momentum of a system of particles depends only on velocity of center of mass (vcm ) of the system.

#### Calculate the momentum of a two particle system as shown. $$(cos \; 37° = 4 /5, \; sin \; 37° = 3 / 5)$$ .

A $$(32\; \hat{i} + 24 \; \hat{j})\; kg\;m/sec$$

B $$(51\; \hat{i} + 20 \; \hat{j})\; kg\;m/sec$$

C $$(24\; \hat{i} + 51 \; \hat{j})\; kg\;m/sec$$

D $$(52\; \hat{i} + 24 \; \hat{j})\; kg\;m/sec$$

×

Write velocities of particle, vectorially

$$m_1 = 4 \; kg, \;\;\;\;\vec{v }_1= 10 \; cos \; 37° + 10 \; sin \; 37 ° = (8\; \hat {i} + 6 \;\hat {j}) \; m/s$$

$$m_2 = 2 \; kg , \;\;\;\;v_2 = 10 \; \hat i \; m/s$$

Calculate velocity of center of mass in x and y– direction, separately.

$$v_{cm(x)} = \dfrac{m_1v_{1x} + m_2v_{2x} +....... + m_n v_{n x}}{\sum \; m}$$

$$v_{cm(y)} = \dfrac{m_1v_{1y} + m_2v_{2y} +....... + m_n v_{n y}}{\sum \; m}$$

Here, $$v_{cm(x)} = \dfrac{4× 8 + 2 × 10 }{4+2} = \dfrac{52}{6} \Rightarrow \dfrac{26}{3} \; m/sec$$

$$v_{cm(y)} = \dfrac{4(6) + 2 (0)}{4+2} = \dfrac{24}{6} = 4 \; m/sec$$

Net momentum is given as

$$\vec {p}_{sys} = \vec{p} _{sys(x)}\; \hat{i} + \vec {p}_{sys(y)}\; \hat{j}$$

$$= \sum m (v_{cm(x)})\hat{i} + \sum m (v_{cm(y)})\hat{j}$$

Here, $$\vec{p}_{sys} = ( 4+2)× \dfrac{26}{3} \hat {i} + (4+2)\times 4 \; \hat {j}$$

$$=( 52 \; \hat {i} + 24 \; \hat {j} )\; kg \; m /sec$$

### Calculate the momentum of a two particle system as shown. $$(cos \; 37° = 4 /5, \; sin \; 37° = 3 / 5)$$ .

A

$$(32\; \hat{i} + 24 \; \hat{j})\; kg\;m/sec$$

.

B

$$(51\; \hat{i} + 20 \; \hat{j})\; kg\;m/sec$$

C

$$(24\; \hat{i} + 51 \; \hat{j})\; kg\;m/sec$$

D

$$(52\; \hat{i} + 24 \; \hat{j})\; kg\;m/sec$$

Option D is Correct

# Momentum of Centre of Mass(COM) of a System

• ## Momentum of system of particles in a system

Consider a n–particle system, having masses $$m_1 , m_2,........m_n$$ with velocities $$\vec{v}_1, \vec{v}_2, .....\vec{v}_n$$ respectively. Momentum  of an individual particle will be given as

For particle 1 : $$\vec{p}_1 = m_1 \, \vec{v}_1$$

For particle 2 : $$\vec{p}_2 = m_2 \, \vec{v}_2$$

For nth particle :  $$\vec{p}_n = m_n \, \vec{v}_n$$

$$m_1 + m_ 2 + .......m_n = \sum m = M$$

• ## Total momentum of all particles

$$\vec{p} = \vec{p}_1 + \vec{p}_2 +......+ \vec{p}_n$$ (Vector sum of individual momentum)

$$\vec{p} = m_1 \vec {v}_1 + m_2 \vec{v}_2 + ... ..+ m_n \vec{v}_n$$

• ## Velocity of center of mass $$(\vec{v}_{cm})$$

Velocity of COM is given by

$$\vec{v}_{cm} = \dfrac{m_1 \vec{v}_1 + m_2 \vec{v }_2 + .... . + m_n \vec{v}_n}{\sum \; m}$$

$$(\sum m) \vec{v}_{cm} = m_1 \vec {v}_1 + m_2 \vec{v}_2 + ..... .+ m_n \vec {v}_n$$

$$M\;\vec{v}_{cm} = \vec {p}_1 + \vec{p}_2 + ...... + \vec {p}_n$$

• Thus, total momentum of system is the vector sum of all the momentum of individual particles which is equal to the momentum of center of mass.

Hence, total momentum of system depends only upon the velocity of COM.

• To understand the concept better, consider an example of a fan. When the fan is rotated each particle on fan has momentum and it is very difficult to calculate total momentum of the system.
• But as we know velocity of COM is zero, so total momentum of fan will be zero, as $$v_{cm} = 0$$.

Note : - Momentum of a system of particles depends only on velocity of center of mass (vcm ) of the system.

#### Calculate the value of v, if total momentum for the given three-particle system is 40 kg m/s in positive x–direction.

A $$- 3.33\; \hat{i} \; m/s$$

B $$6.66 \; \hat{i} \; m/s$$

C $$10.44 \; \hat{i} \; m/s$$

D $$11.50 \; \hat{i} \; m/s$$

×

List the given velocities and momentum vectorially

$$m_1 = 4\; kg , \;\;\;\vec{v}_1 = 10 \; \hat{i} \; m/s$$

$$m_2 = 2\; kg , \;\;\;\vec{v}_2 = 5 \; \hat{i} \; m/s$$

$$m_3 = 3\; kg , \;\;\;\vec{v}_3 = v\; m/s$$

$$\vec{p} _{sys} = 40 \; kg - m/s$$

Momentum of particles ( of a system) is given by

$$\vec{p}_{sys} = (m_1 + m_2 + ..... + m_n) v_{cm}$$

$$\vec{p}_{sys} = m_1 \vec{v}_1 + m_2 \vec{v}_2 + ..... + m_n \vec{v}_n$$

Substituting values

$$40 \,\hat{i} = 4 × (10 \, \hat{i}) + 2(5 \, \hat{i}) + 3 (v)$$

$$\Rightarrow 3v = - 10 \, \hat{i}$$

$$\Rightarrow v = -\left( \dfrac{10}{3}\right)\; \hat{i} \; m/s$$

$$\Rightarrow v = - 3. 33 \; \hat{i} \; m/s$$

Thus, velocity, $$v = - 3. 33 \; \hat{i} \; m/s$$

### Calculate the value of v, if total momentum for the given three-particle system is 40 kg m/s in positive x–direction.

A

$$- 3.33\; \hat{i} \; m/s$$

.

B

$$6.66 \; \hat{i} \; m/s$$

C

$$10.44 \; \hat{i} \; m/s$$

D

$$11.50 \; \hat{i} \; m/s$$

Option A is Correct

# Acceleration of Center of Mass for Atwood Machine(1 D)

• ## Center of mass for n–particle system

Position of center of mass for n–particle system is given by

$$\vec{r}_{cm} = \dfrac{m_1 \vec{r_1} + m_2\vec{r_2} + m_3\vec{r_3} + ......+ m_n \vec{r_n}}{m_1 +m_2 +m_3......+ m_n}$$

Here, $$m_1,m_2.....m_n$$ are masses of particles and $$\vec{r_1}, \;\;\vec{r_2},.....\vec{r_n}$$ are their respective position vectors.

or,

$$\vec{r_{cm }} = \dfrac{m_1\vec{r_1} + m_2\vec{r_2} + ....+ m_n\vec{r_n}}{\sum\limits_{i = 1}^{ n} \; m_i}\;\;\;\;\;\;\;\;\;— (1)$$

• ## Velocity of center of mass

From equation (1)

$$\sum \; m_i (\vec{r_{cm}}) = m_1 \vec{r_1} + m_2 \vec{r_2} + ......+m_n \vec{r_n}$$

Differentiating the above equation on both sides

$$\sum \; mi (\vec{v}_{cm}) = m_1 \vec {v}_1 + m_2 \vec {v}_2 + .....m_n \vec {r}_n\;\;\;\;\;— (2)$$

Thus,   $$\vec{v}_{cm} = \dfrac{m_1 \vec {v}_1 + m_2 \vec {v}_2 + .... + m_n \vec {r}_n}{\sum\limits_{i= 1}^{n} \; m_i}\;\;\;\;\;— (3)$$?

• ## Acceleration of Center of Mass

Again differentiating equation (2)

$$\sum m_i (\vec{a}_{cm}) = m_1\vec{a_1} + m_2\vec{a_2} + ..... +m_n\vec{a_n}$$

or,

$$\vec{a_{cm}} = \dfrac{m_1\vec{a_1} + m_2\vec{a_2} + .... m_n\vec{a_n}}{\sum\limits_{i = 1} ^{ n} \; m_i}\;\;\;\;\;\;\;\;—(4)$$

#### For the Atwood - machine, as shown in figure, find the acceleration of center of mass.

A $$\dfrac{ -5}{9 }\; \hat{j} \; m/s^2$$

B $$\dfrac{-10}{ 9}\; \hat{j} \; m/s^2$$

C $$\dfrac{ -15}{9 }\; \hat{j} \; m/s^2$$

D $$\dfrac{-12 }{9 } \; \hat{j} \; m/s^2$$

×

Find velocity/acceleration of individual masses in system.

Taking upward motion along + y–axis.

Acceleration (a) of masses $$= \dfrac{(m_2 - m_1)}{m_1 + m_2}g\;\;\;\; [m_2 >m_1]$$

$$\therefore$$ Taking, $$m_1 = 10 \; kg , \;\;\; g = 10 \; m/s^2, \;\;\;\;\;m_2 = 20 \; kg$$

$$a = \dfrac{[20 -10 ]}{30} × 10 \;\; =\dfrac{10}{3}\; m/s^2$$

List the given velocities/accelerations vectorially, for each particle.

For $$m_1 = 10 \; kg\; \; \Rightarrow \vec {a}_1 = \dfrac{10}{3} \; \hat{j} \; m/s^2$$

$$m_2 = 20 \; kg \; \Rightarrow \;\; \vec {a}_2 = \dfrac{ -10}{3}\; \hat{j} \; m/s^2$$

Using equation for acceleration of center of mass

$$\vec{a}_{cm}= \dfrac{m_1 \vec{a}_1 +m_2\vec{a}_2}{m_1 + m_2}$$

$$\vec{a}_{cm} = \dfrac{10 × \dfrac{10}{3} \; \hat {j} + 20 \left(\dfrac{-10}{3}\right)\; \hat{j}}{30}$$

$$\vec{a}_{cm} = \dfrac{-100}{3 × 30}\; \hat{j} = \dfrac{-10}{9}\; \hat{j} \; m/s^2$$

### For the Atwood - machine, as shown in figure, find the acceleration of center of mass.

A

$$\dfrac{ -5}{9 }\; \hat{j} \; m/s^2$$

.

B

$$\dfrac{-10}{ 9}\; \hat{j} \; m/s^2$$

C

$$\dfrac{ -15}{9 }\; \hat{j} \; m/s^2$$

D

$$\dfrac{-12 }{9 } \; \hat{j} \; m/s^2$$

Option B is Correct