Question 249: Centre of Mass and Rotational Motion

Question

Three identical spheres each of mass $2 \mathrm{M}$ are placed at the corners of a right angled triangle with mutually perpendicular sides equal to $4 \mathrm{~m}$ each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is $\frac{4 \sqrt{2}}{x}$, where the value of $x$ is ___________ .

Accepted Answer

1. Center of Mass Coordinates:
The center of mass (COM) of a system of particles is calculated as:
$X_{COM} = \frac{\sum_{i} m_i x_i}{\sum_{i} m_i}$
$Y_{COM} = \frac{\sum_{i} m_i y_i}{\sum_{i} m_i} $$$
where $m_i$ is the mass of the $i$-th particle, and $(x_i, y_i)$ are its coordinates.
2. Coordinate Setup:
Let's place the origin at the right angle of the triangle and align the sides along the x and y axes:
Sphere 1: $(4, 0)$
Sphere 2: $(0, 4)$
Sphere 3: $(0, 0)$
3. Calculations:
Since all spheres have mass $2M$, we can simplify the COM calculations:
$$X_{COM} = \frac{2M \cdot 4 + 2M \cdot 0 + 2M \cdot 0} {2M + 2M + 2M} = \frac{4}{3}$$
$$Y_{COM} = \frac{2M \cdot 0 + 2M \cdot 4 + 2M \cdot 0} {2M + 2M + 2M} = \frac{4}{3}$$
4. Magnitude of Position Vector:
The position vector of the COM is $\left(\frac{4}{3}, \frac{4}{3}\right)$.  Its magnitude is:
$$|\vec{r}_{COM}| = \sqrt{\left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2} = \frac{4\sqrt{2}}{3}$$
5. Finding x:
We are given that the magnitude of the position vector is of the form $\frac{4\sqrt{2}}{x}$. Comparing this to our result, we find that $x = 3$.
Answer:
The value of $x$ is 3.

Centre of Mass and Rotational Motion

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