Question

# a point on the rim of a rotating wheel 4 m in diameter has a velocity of 1600 cm/s. the angular velocity of the wheel is

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Solution

## Dear Student, $\mathrm{linear}\mathrm{velocity},\mathrm{v}=1600\mathrm{cm}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{or},\mathrm{v}=16\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{diameter},\mathrm{d}=4\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{radius},\mathrm{r}=\frac{\mathrm{d}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{or},\mathrm{r}=\frac{4}{2}=2\mathrm{m}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know}\mathrm{that},\phantom{\rule{0ex}{0ex}}\mathrm{v}=\mathrm{r\omega }\phantom{\rule{0ex}{0ex}}\mathrm{where},\mathrm{\omega }=\mathrm{angular}\mathrm{velocity}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}16=2×\mathrm{\omega }\phantom{\rule{0ex}{0ex}}\mathrm{\omega }=\frac{16}{2}=8\mathrm{rad}/\mathrm{s}$

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