Question

A wheel is of diameter $1\mathrm{m}$. If it makes $30\mathrm{revolutions}/\sec$, then the linear speed (in m/sec) of a point on its circumference is:

• A. $30\mathrm{\pi }$
• B. $\mathrm{\pi }$
• C. $60\mathrm{\pi }$
• D. $\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Answer 1

The correct option is A

$30\mathrm{\pi }$

Step 1: Given data:

The diameter of the wheel($\mathrm{d}$) = $1\mathrm{m}$

Thus, the radius($\mathrm{r}$) of the wheel = $\frac{\mathrm{Diameter}}{2}=\frac{1\mathrm{m}}{2}=0.5\mathrm{m}$

Angular speed($\mathrm{\omega }$) of the wheel = $30\mathrm{revolutions}/\sec$

since in 1 complete revolution the angular displacement($\mathrm{\theta }$)= $2\mathrm{\pi }\mathrm{radian}\left(\mathrm{rad}\right)$

therefore the angular speed($\mathrm{\omega }$) of the wheel = $30\times 2\mathrm{\pi }\mathrm{radian}/\sec$

Step 2: Calculation of linear speed($\mathrm{v}$) :

$$\begin{gathered} \mathrm{Linear}\mathrm{speed}\left(\mathrm{v}\right)=\mathrm{radius}\left(\mathrm{r}\right)\times \mathrm{angular}\mathrm{speed}\left(\mathrm{\omega }\right) \\ \mathrm{v}=\frac{1}{2}\mathrm{m}\times 30\times 2\mathrm{\pi rad}/\sec \\ \mathrm{v}=30\mathrm{\pi }\mathrm{m}{\mathrm{s}}^{-1} \end{gathered}$$

Thus,

Option a) $30\mathrm{\pi }$ is the correct option.