A dentist using a dental drill brings it from rest to maximum operating speed of 382,000 rpm in 3.0 s. Assume that the drill accelerates at

Question

A dentist using a dental drill brings it from rest to maximum operating speed of 382,000 rpm in 3.0 s. Assume that the drill accelerates at a constant rate during this time.
A) What is the angular acceleration of the drill in rev/s2?
B) Find the number of revolutions the drill bit makes during the 2.6 s time interval. rev

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Thành Công 5 years 2021-08-15T21:11:44+00:00 1 Answers 33 views 0

Answers ( )

  1. thienthanh
    0
    2021-08-15T21:13:14+00:00
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    Answer:

    a. \alpha =2122.22\: rev/s^{2}

    b. \Delta \theta =9,550.02\: rev

    Explanation:

    The computation is shown below:

    data provided in the question

    The initial angular velocity  \small \omega o = 0 rev/s,

    f = Final angular velocity = \small \omega = 382000 rpm i.e = \frac{382,000}{60} = 6,366.67

    And time = t = 3.0s

    Based on the above information

    a. For angular acceleration of drill

    \small \omega =\omega _{o}+\alpha t \\\\\ 6,366.67 = 0 + \alpha (3.0) \\\\ \alpha =2122.22\: rev/s^{2}

    b. For the number of revolutions

    \small \omega ^{2}-\omega _{o}^{2}=2\alpha \Delta \theta \\\\(6,366.67) ^{2}-(0)^{2}=2(2122.22) \Delta \theta \\\\ \Delta \theta =9,550.02\: rev

    We simply applied the above formulas for determining each parts

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