Question

A heavy particle of mass m falls freely near the earth's surface. What is the torque acting on this particle about a point 50 cm east to the line of motion? Does this torque produce any angular acceleration in the particle?

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}: \\ \overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F} \\ \mathrm{Given}: \\ \overrightarrow{r}=-0.5\hat{\mathrm{i}}\mathrm{m} \\ \overrightarrow{F}=-\mathrm{mg}\hat{\mathrm{j}} \\ \mathrm{The}\mathrm{torque}\mathrm{becomes}: \\ \overrightarrow{\tau }=0.5\left(-\hat{\mathrm{i}}\right)\times mg\left(-\hat{\mathrm{j}}\right) \\ \overrightarrow{\tau }=0.5\mathrm{mg}\hat{\mathrm{k}}\left[\because \hat{\mathrm{i}}\times \hat{\mathrm{j}}=\hat{\mathrm{k}}\right] \\ \mathrm{No},\mathrm{there}\mathrm{will}\mathrm{be}\mathrm{no}\mathrm{angular}\mathrm{acceleration}\mathrm{on}\mathrm{the}\mathrm{particle}\mathrm{due}\mathrm{to}\mathrm{the}\mathrm{torque}. \\ \mathrm{Angular}\mathrm{acceleration}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{\alpha }=\frac{\tau }{I}.\mathrm{As}\mathrm{the}\mathrm{particle}\mathrm{here}\mathrm{moves}\mathrm{in}\mathrm{a} \\ \mathrm{straight}\mathrm{line},\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{rotation}\mathrm{lies}\mathrm{at}\mathrm{a}\mathrm{distance}\mathrm{infinity}(r=\infty );\mathrm{so},\mathrm{moment}\mathrm{of}\mathrm{inertia}(I=m{r}^2)\mathrm{of}\mathrm{the} \\ \mathrm{particle}\mathrm{is}\mathrm{infinity}. \\ \therefore \alpha =0 \end{gathered}$$