Question 319: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

$$\begin{aligned}
&	ext { For equilibrium, }\
&\begin{aligned}
& N_1 r=W(r-x) \
	herefore \quad & N_1=\frac{W(r-x)}{r}
\end{aligned}
\end{aligned}$$

Answer

$\frac{\mathrm{W} \cdot \mathrm{r}}{\mathrm{x}}$

Answer

$\frac{\mathrm{W} \cdot \mathrm{x}}{\mathrm{r}}$

Answer

$\mathrm{\frac{W \cdot(r-x)}{x}}$

Answer

$\mathrm{\frac{W \cdot(r-x)}{r}}$

Centre of Mass and Rotational Motion

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

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