A wheel of radius 1 m rolls through $180^{\circ}$ over a plane surface. The magnitude of the displacement of the point of the wheel initially in contact with the surface is.

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Aicharya · Accepted Answer

Distance travelled by the wheel in half revolution $=\frac{\mathrm{C}}{2}=\frac{2 \pi \mathrm{r}}{2}=\pi \mathrm{r}$ Where C is the circumference of the wheel. $\therefore \quad$ From figure, Displacement of initial point of contact after half revolution $=\mathrm{AB}$ $\begin{array}{ll} \therefore \quad & A B^2=A D^2+\mathrm{DB}^2 \\ & A B^2=(\pi r)^2+(2 r)^2=r^2\left(\pi^2+4\right) \\ \therefore \quad & A B=r \sqrt{\left(\pi^2+4\right)} \\ \therefore \quad & A B=\sqrt{\left(\pi^2+4\right)} \end{array}$ ...(given $r=1 \mathrm{~m}$ )

A wheel of radius 1 m rolls through $180^{\circ}$ over a plane surface. The magnitude of the displacement of…

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