ROTATIONAL MOTION

1 )

The centre of mass of systems of two particles is

on the line joining them and midway between them

on the line joining them at a point whose distance from each particle is proportional to the square of the mass of that particle.

on the line joining them at a point whose distance from each particle inversely proportional to the mass of that particle.

On the line joining them at a point whose distance from each particle is proportional to the mass of that particle.

View Solution

Answer :

on the line joining them at a point whose distance from each particle inversely proportional to the mass of that particle.

For two particle system,

R_CM = [(m₁r₁ + m₂r₂) / (m₂ + m₂)]

if measurement is done from centre of mass point then

m₁r₁ = m₂r₂

(m₁ / m₂) = (r₂ / r₁) i.e m ∝ (1/r) i.e r ∝ (1/m)

Hence centre of mass is on line joining them and distance is inversely proportional to mass

on the line joining them at a point whose distance from each particle inversely proportional to the mass of that particle.

For two particle system,

R_CM = [(m₁r₁ + m₂r₂) / (m₂ + m₂)]

if measurement is done from centre of mass point then

m₁r₁ = m₂r₂

(m₁ / m₂) = (r₂ / r₁) i.e m ∝ (1/r) i.e r ∝ (1/m)

Hence centre of mass is on line joining them and distance is inversely proportional to mass

2 )

Particles of 1 gm, 1 gm, 2 gm, 2 gm are placed at the corners A, B, C, D, respectively of a square of side 6 cm as shown in figure. Find the distance of centre of mass of the system from geometrical centre of square.

1 cm

2 cm

3 cm

4 cm

View Solution

Answer :

from diagram, geometric centre of square is P(3, 3).

r_A = 0i + 6j

r_B = 6i + 6j

r_C = 6i + 0j

r_D = 0i + 0j

R = [(m_Ar_A + m_Br_B + m_Cr_C + m_Dr_D) / (m_A + m_B + m_C + m_D)]

= [{(1)(6j˄) + (1)(6i˄ + 6j˄) + (2)(6i˄) + (2)(0)} / {1 + 1 + 2 + 2}]

R = [(12j˄ + 18i˄) / 6] = 3i˄ + 2j˄

P = 3i˄ + 3j˄

distance between point P and R = √[(3 – 3)² + (3 – 2)²]

= 1 cm

3 )

Three particles of the same mass lie in the (X, Y) plane, The (X, Y) coordinates of their positions are (1, 1), (2, 2) and (3, 3) respectively. The (X, Y) coordinates of the centre of mass are

(1, 2)

(2, 2)

(1.5, 2)

(2, 1.5)

View Solution

Answer :

x = [(m₁x₁ + m₂x₂ + m₃x₃) / (m₁ + m₂ + m₃)]

and y = [(m₁y₁ + m₂y₂ + m₃y₃) / (m₁ + m₂ + m₃)]

given m₁ = m₂ = m₃ = m(say)

also (x₁, y₁) = (1, 1)

(x₂, y₂) = (2, 2)

(x₃, y₃) = (3, 3)

x = (1/3)(1 + 2 + 3)

= 2

y = (1/3)(1 + 2 + 3)

= 2

centre of mass = (x, y) = (2, 2)

4 )

Consider a two-particle system with the particles having masses M₁, and M₂. If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved so as to keep the centre of mass at the same position?

Select correct option from the following options

{A} [(M₁d) / (M₁ + M₂)] {B} [(M₂d) / (M₁ + M₂)]

{C} [(M₁d) / (M₂)] {D} [(M₂d) / (M₁)]

View Solution

Answer :

m₁x₁ = m₂x₂

after movement of m₁ by d & m₂ by d'

m₁(x₁ – d) = m₂(x₂ – d')

m₁x₁ – m₁d = m₂x₂ – m₂d'

∴ (m₁/m₂) = (d'/d)

d' = [(m₁ / m₂)] ∙ d

5 )

Four particles A, B, C and D of masses m, 2m, 3m and 4m respectively are placed at corners of a square of side x as shown in figure find the coordinate of centre of mass take A at origin of x-y plane.

[2x, (7x / 10)]

[2x, (10x / 7)]

[(x / 2), (10x / 7)]

[(x / 2), (7x / 10)]

View Solution

Answer :

m₁ = m, m₂ = 2m, m₃ = 3m, m₄ = 4m

A(0, 0), B(x, 0), C(x, x), D(0, x),

x_CM = [(m₁x₁ + m₂x₂ + m₃x₃ + m₄x₄) / (m₁ +m₂ + m₃ + m₄)]

= [(m × 0) + (2mx) + (3mx) + (4m × 0)} / {m + 2m + 3m + 4m}]

= [(5mx) / (10m)] = (x/2)

Y_CM = [(m₁y₁ + m₂y₂ + m₃y₃ + m₄y₄) / (m₁ + m₂ + m₃ + m₄)]

= [(m × 0) + (2m × 0) + (3mx) + (4mx)} / {10m}]

= [(7mx) / (10m)]

= (7 / 10)x

(x, y) = [(x/2), (7x / 10)]

Looking at options given,

centre of mass of system has to be less than distance x.

hence option D only satisfies this or criteria.

Topic Categories