Physics, 01.08.2019 23:30 BigCra

Asafety device brings the blade of a power mower from an initial angular speed of w1 to rest in 5.00 revolutions. at the same constant accelerations, how many revolutions would it take the blade to come to rest from an initial angular speed w3 that was three times as great, w3 = 3w1?

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We have that the revolutions would it take the blade to come to rest from an initial angular speed w3 that was three times as great

n_2=45rev

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A safety device brings the blade of a power mower from an initial angular speed of w1 to rest in 5.00 revolutions. at the same constant accelerations, how many revolutions would it take the blade to come to rest from an initial angular speed w3 that was three times as great, w3 = 3w1?

Generally the equation for the Revolutions before rest is mathematically given as

$For n_1=5\\\\w_1^2=\w_1^2+2\aplhan_1\\\\\alpha=\frac{-w_1^2}{10}$

For n_2

$w_4^2=w_3^2+2\alphan_2\\\\\frac{w_1^2n_2}{5}=9w_1^2\\\\\n_2=45rev$

Therefore

The revolutions would it take the blade to come to rest from an initial angular speed w3 that was three times as great

n_2=45rev

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Answer from: carolina1636

Hello!

If the acceleration is a constant omega prime is 3times omega inital then it would take 15revs to stop. 3 times the revolutions because everything is held as a constant and the only change is omega.

Theta final=Theta initial + omega initial * t +1/2at^2

Theta is revs and this is what you want to find. so lets set it up and check. assuming everything stays constant BUT omega, then we can treat everything else in the equation as a one, or essentially disregard them. So we see a linear relationship to revolutions or theta, meaning whatever factor omega is increased or decreased then if everything else is a constant then revolutions will change by the same factor. Omega is angular speed or how fast something is spinning on an axis of rotation.

Hope this explanation helps. Any questions please ask! Thank you so much!

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