G i v e n m A = m , m B = 2 m , m C = 3 m Given\ m_{A} = m,m_{B} = 2m,m_{C} = 3m a n d m D = 4 m … … ( i ) and\ \ m_{D} = 4m\ldots\ldots(i) a A = − a i ̂ , a B = + a j ̂ , a C = + a i ̂ a_{A} = - a\widehat{i},a_{B} = + a\widehat{j},a_{C} = + a\widehat{i} a n d a D = − a j ̂ … … ( i i ) and\ a_{D} = - a\widehat{j}\ldots\ldots(ii) To find: Acceleration of centre of mass of the particles, a C M a_{CM} .Acceleration of centre of mass for a system of masses is given as: a ⃗ c m = m 1 a ⃗ 1 + m 2 a ⃗ 2 + m 3 a ⃗ 3 + m 4 a ⃗ 4 m 1 + m 2 + m 2 + m 4 {\overrightarrow{a}}_{cm} = \frac{m_{1}{\overrightarrow{a}}_{1} + m_{2}{\overrightarrow{a}}_{2} + m_{3}{\overrightarrow{a}}_{3} + m_{4}{\overrightarrow{a}}_{4}}{m_{1} + m_{2} + m_{2} + m_{4}} = m ( − a i ̂ ) + 2 m ( a j ̂ ) + 3 m ( a i ̂ ) + 4 m ( − a j ̂ ) m + 2 m + 3 m + 4 m = \frac{m( - a\widehat{i}) + 2m(a\widehat{j}) + 3m(\widehat{ai}) + 4m( - a\widehat{j})}{m + 2m + 3m + 4m} a ⃗ c m = a ( 2 i ̂ − 2 j ̂ ) 10 = a 5 ( i ̂ − j ̂ ) {\overrightarrow{a}}_{cm} = \frac{a(2\widehat{i} - 2\widehat{j})}{10} = \frac{a}{5}(\widehat{i} - \widehat{j})

Four particles A, B, C and D with masses mA = m, mB = 2m, mC = 3m and mD = 4m are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the centre of mass of the particles is

Question 1:

Four particles A, B, C and D with masses m_A = m, m_B = 2m, m_C = 3m and m_D = 4m are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the centre of mass of the particles is

$\frac{a}{5}(\widehat{i} + \widehat{j})$

$a(\widehat{i} + \widehat{j})$

zero

$\frac{a}{5}(\widehat{i} - \widehat{j})$

Answer:

Solution:

$Given\ m_{A} = m,m_{B} = 2m,m_{C} = 3m$

$and\ \ m_{D} = 4m\ldots\ldots(i)$

$a_{A} = - a\widehat{i},a_{B} = + a\widehat{j},a_{C} = + a\widehat{i}$

$and\ a_{D} = - a\widehat{j}\ldots\ldots(ii)$

To find: Acceleration of centre of mass of the particles, $a_{CM}$ .

Acceleration of centre of mass for a system of masses is given as:

${\overrightarrow{a}}_{cm} = \frac{m_{1}{\overrightarrow{a}}_{1} + m_{2}{\overrightarrow{a}}_{2} + m_{3}{\overrightarrow{a}}_{3} + m_{4}{\overrightarrow{a}}_{4}}{m_{1} + m_{2} + m_{2} + m_{4}}$

$= \frac{m( - a\widehat{i}) + 2m(a\widehat{j}) + 3m(\widehat{ai}) + 4m( - a\widehat{j})}{m + 2m + 3m + 4m}$

${\overrightarrow{a}}_{cm} = \frac{a(2\widehat{i} - 2\widehat{j})}{10} = \frac{a}{5}(\widehat{i} - \widehat{j})$

Prev Exam(s):

JEE MAIN

- 2019

, 8 Apr Shift-01

Chapter:

CENTER OF MASS (COM)

Question Level:

Tags:

CENTER OF MASS (COM)
PHYSICS