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24 July, 03:57

5. A 1.15-kg grinding wheel 22.0 cm in diameter is spinning counterclockwise at a rate of 20.0 revolutions per second. When the power to the grinder is turned off, the grinding wheel slows with constant angular acceleration and takes 80.0 s to come to a rest? (a) What was the angular acceleration (in rad/s2) of the grinding wheel as it came to rest if we take a counterclockwise rotation as positive? (b) How many revolutions did the wheel make during the time it was coming to rest?

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  1. 24 July, 04:14
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    a) - 1.57 rad/s²

    b) 800 revolutions

    Explanation:

    a) In here you can use the equations of velocity as if it were a linear movement. In this case:

    wf = wo + at

    wo is the innitial angular velocity, that we can get this value using the fact that a revolution is 2π so:

    wo = 20 * 2π = 125.66 rad/s

    We have the time of 80 seconds, and the final angular speed is zero, because it's going to a rest so:

    0 = 125.66 + 80a

    a = - 125.66 / 80

    a = - 1.57 rad/s²

    b) In this part, we will use the following expression:

    Ф = Фo + wo*t + 1/2 at²

    But as this it's coming to rest then:

    Ф = 1/2at²

    solving we have:

    Ф = 0.5 * (-1.57) * (80) ²

    Ф = 5,024 rad

    Ф = 5024 / 2π

    Ф = 800 revolutions
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