Question

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 ______.

विकल्प

• `a/5(hati - hatj)`

• `a(hati + hatj)`

• Zero

• `a/5(hati + hatj)`

Answer 1

Four particles A, B, C, and D with masses m_A = m, m_B = 2 m, 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 `underlinebb(a/5(hati - hatj))`.

Explanation:

Given: m_A = m, m_B = 2m, m_C = 3m, and m_D = 4m ........(i)

`a_A = -ahati, a_B = +ahatj, a_C = +ahati  "and"  a_D = -ahatj` ............(ii)

To find: The acceleration of the particle's centre of mass a_CM.

For a system of masses, the acceleration of the centre of mass is given as:

`a_{CM} = (m_Aa_A + m_Ba_B + m_Ca_C + m_Da_D)/(m_A + m_B + m_C + m_D)` ......(iii)

Putting values from equations (i) and (ii) in equation (iii),

`a_{CM} = (-mahati + 2mahatj + 3mahati - 4mahatj)/(m + 2m + 3m + 4m)`

= `(2mahati - 2mahatj)/(10m)`

`a_{CM} = a/5(hati - hatj)`