Question

Three thin walled uniform

hollow spheres of radii 1cm, 2 cm and 3 cm

are so located that their centres are on the three

vertices of an equilateral triangle ABC having

each side 10 cm. Determine centre of mass of

the system​

Answer 1

Answer:

answer of your question

Answer 2

Answer:

The center of mass of the system lies at ( $\frac{45}{14}$,$\frac{5\sqrt{3} }{14}$).

Explanation:

We will solve this question through the following relation,

$x_c_m=\frac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3}$       (1)

$y_c_m=\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}$      (1)

Where,

$x_c_m$= center of mass on x-coordinate

$y_c_m$=center of mass on y-coordinate

m₁,m₂,m₃=respective masses

x₁,x₂,x₃=respective position of masses on x-coordinate

y₁,y₂,y₃=respective position of masses on y-coordinate

The mass of the thin-walled uniform hollow sphere is proportional to its surface area. So,

m₁=m

m₂=4m

m₃=9m

The coordinates of mass m₁=5,$5\sqrt{3}$

The coordinates of mass m₂=10,0

The coordinates of mass m₃=0,0

By substituting the required values in equation (1) we get;

$x_c_m=\frac{m\times 5+4m\times 10+9m\times 0}{m+4m+9m}=\frac{45}{14}cm$

$y_c_m=\frac{m\times 5\sqrt{3} +4m\times 0+9m\times 0}{m+4m+9m}=\frac{5\sqrt{3} }{14} cm$

Hence, the center of mass of the system lies at ( $\frac{45}{14}$,$\frac{5\sqrt{3} }{14}$).