A rigid body in the X–Y plane consists of two point masses (1 kg each) attached to the ends of two massless rods, each of 1 cm length, as shown in the figure. It rotates at 30 RPM counter-clockwise about the Z-axis passing through point O. A point mass of $\sqrt{2$ kg, attached to one end of a third massless rod, is used for balancing the body by attaching the free end of the rod to point O. The length of the third rod is ________________ cm.}
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Dynamic balancing of rotating systems requires the vector sum of all $m r$ terms to be zero. Always treat unbalance as a vector quantity.
Since the system rotates about O, the rotating masses must be dynamically balanced. Dynamic balance requires the vector sum of the rotating mass moments to be zero:
\[
\sum m_i r_i = 0.
\] Step 1: Identify the mass positions.
Two masses of 1 kg each are located at:
- One at 1 cm along the positive X-axis
- One at 1 cm rotated by 90° (i.e., along the positive Y-axis)
Thus their position vectors are:
\[
\mathbf{r}_1 = 1\,\hat{i}, \qquad \mathbf{r}_2 = 1\,\hat{j}.
\] Step 2: Compute their combined unbalance vector.
\[
\mathbf{U} = 1(1\,\hat{i}) + 1(1\,\hat{j}) = \hat{i} + \hat{j}.
\]
This vector has magnitude:
\[
|\mathbf{U}| = \sqrt{1^2 + 1^2} = \sqrt{2}.
\] Step 3: Balance using a third mass of $\sqrt{2$ kg.}
For perfect balance,
\[
m_3 r_3 = |\mathbf{U}|.
\]
Given:
\[
m_3 = \sqrt{2}.
\]
Thus,
\[
\sqrt{2}\, r_3 = \sqrt{2} \Rightarrow r_3 = 1 \text{ cm}.
\]
Therefore, the length of the third rod must be 1 cm. Final Answer: 1 cm
