A $1.5$-foot radius bicycle wheel is spun at a speed of $4$ radians per second. How fast is a point on the edge of the wheel moving in feet per second?
Would I just do the following?:
$\frac{4\space\text{rad}}{2\pi}\times2\pi(1.5) $
$1.5$-foot radius bicycle wheel spun at $4$ radians per second; how fast is a point on the edge of the wheel moving?
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algebra-precalculus
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0$v=\omega\cdot r$, so it seems right. – 2017-01-25
1 Answers
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As MattG88 pointed out, $v=\omega \cdot r$, where $v$ is the tangential velocity, $\omega$ is the rotational velocity, and $r$ is the radius.
However, in there is a mistake in your equation. You should have
$\omega =\frac{4}{2\pi}$
$r=1.5$
because $\omega$ is the rate of change in (radian) angle per second, and $r$ is the radius.