Three point masses 1.0kg , 1.5kg and 2.5kg are placed at the vertices of a right-angle triangle of sides 4.0cm , 3.0cm and 5.0cm , as shown in the figure. The centre of mass of the system is

Question

Three point masses 1.0kg , 1.5kg and 2.5kg are placed at the vertices of a right-angle triangle of sides 4.0cm , 3.0cm and 5.0cm , as shown in the figure. The centre of mass of the system is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-uyd1ciofktzxzohr.jpg" /> (A) 0.6cm right and 2.0cm above 1kg mass (B) 2.0cm right and 0.9cm above 1kg mass (C) 0.9cm right and 2.0cm above 1kg mass (D) 1.5cm right and 1.2cm above 1kg mass

Tardigrade Admin · Accepted Answer

Let us assume that the origin of the coordinate system is at the location of

1 k g

mass, then
Solution

x_{c m} = \frac{1 \times 0 + 1.5 \times 3 + 2.5 \times 0}{5} = 0.9 c m

y_{c m} = \frac{1 \times 0 + 1.5 \times 0 + 2.5 \times 4}{5} = 2 c m

Q. Three point masses $1.0 k g$ , $1.5 k g$ and $2.5 k g$ are placed at the vertices of a right-angle triangle of sides $4.0 c m$ , $3.0 c m$ and $5.0 c m$ , as shown in the figure. The centre of mass of the system is

$0.6 c m$ right and $2.0 c m$ above $1 k g$ mass

$2.0 c m$ right and $0.9 c m$ above $1 k g$ mass

$0.9 c m$ right and $2.0 c m$ above $1 k g$ mass

$1.5 c m$ right and $1.2 c m$ above $1 k g$ mass

Let us assume that the origin of the coordinate system is at the location of $1 k g$ mass, then

$x_{c m} = \frac{1 \times 0 + 1.5 \times 3 + 2.5 \times 0}{5} = 0.9 c m$
$y_{c m} = \frac{1 \times 0 + 1.5 \times 0 + 2.5 \times 4}{5} = 2 c m$

Q. Three point masses 1.0kg , 1.5kg and 2.5kg are placed at the vertices of a right-angle triangle of sides 4.0cm , 3.0cm and 5.0cm , as shown in the figure. The centre of mass of the system is

0.6cm right and 2.0cm above 1kg mass

2.0cm right and 0.9cm above 1kg mass

0.9cm right and 2.0cm above 1kg mass

1.5cm right and 1.2cm above 1kg mass