MHT CET · Physics · Center of Mass Momentum and Collision
A metal rod of weight ' \(W\) ' is supported by two parallel knife-edges \(A\) and \(B\). The rod is. in equilibrium in horizontal position. The distance between two knife-edges is ' \(r\) '. The centre of mass of the rod is at a distance ' \(x\) ' from A. The normal reaction on A is
- A \(\frac{\mathrm{W} \cdot \mathrm{r}}{\mathrm{x}}\)
- B \(\frac{\mathrm{W} \cdot \mathrm{x}}{\mathrm{r}}\)
- C \(\frac{W \cdot(r-x)}{x}\)
- D \(\frac{\mathrm{W} \cdot(\mathrm{r}-\mathrm{x})}{\mathrm{r}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{W} \cdot(\mathrm{r}-\mathrm{x})}{\mathrm{r}}\)
Step-by-step Solution
Detailed explanation

For equilibrium,
\(\begin{aligned}& \mathrm{N}_1 \mathrm{r}=\mathrm{W}(\mathrm{r}-\mathrm{x}) \\\therefore \quad & \mathrm{N}_1=\frac{\mathrm{W}(\mathrm{r}-\mathrm{x})}{\mathrm{r}}\end{aligned}\)
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