A metre stick is held vertically with one end of it on the floor and is then allowed to fall. The sp

Question

A metre stick is held vertically with one end of it on the floor and is then allowed to fall. The speed of the other end when it hits the floor, assuming that the end of the stick does not slip and $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}$\

Professor PC Thomas Classes · Answer

\In the vertical position, the metre stick has got $\mathrm{PE}=\mathrm{m}$ gh. When it falls, it is converted into rotational $\mathrm{KE}=\frac{1}{2} \mathrm{I} \omega^{2}$.\ $	herefore \frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{mgh}$\ $\frac{1}{2}\left(\frac{\mathrm{ml}{ }^{2}}{3}ight) \omega^{2}=\mathrm{mg}\left(\frac{l}{2}ight)$\ Calculate $\omega$\ $\omega$ is the same at all points. But v is different for different points because in the equation $\mathrm{v}=\mathrm{r} \omega$; r is different at different points. Here $\mathrm{r}=1 \mathrm{~m}$.\ $	herefore \mathrm{v}=$\