Question #7c8db

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Question #7c8db

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Answer 1 · 2017-10-04T22:46:37

Whether you use $a_{c} = \frac{v^{2}}{r}$ $a_c=v^2/r$ or $a_{c} = ω^{2} r$ $a_c=omega^2r$ depends on the manner in which the velocity or acceleration is stated.
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Explanation:

In this problem, the acceleration is stated in ${metres per second}^{2}$ $"metres per second"^2$ . This determines the units and also the form of the relation you must use, namely:

$a_{c} = \frac{v^{2}}{r}$ $a_c=v^2/r$ f

Had the acceleration been stated in ${radians per second}^{2}$ $"radians per second"^2$ , you would use the $a_{c} = ω^{2} r$ $a_c=omega^2r$ form.

Both formulas would yield the same result, of course, because $v$ $v$ increases as $r$ $r$ increases, but $ω$ $omega$ does not.

In other words, when $r$ $r$ doubles, $v^{2}$ $v^2$ increases by $2^{2}$ $2^2$ times, and $\frac{v^{2}}{r}$ $v^2/r$ increases by two times.

When $r$ $r$ doubles, $ω$ $omega$ does not change, so $ω^{2} r$ $omega^2r$ doubles.

Finally, in the last part of the question, if $ω$ $omega$ is doubled, then $ω^{2} r$ $omega^2r$ increases by $2^{2}$ $2^2$ .

Hope that clears it up.

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