A record player rotates a record at 45 revolutions per minute. When the record player is switched off, it makes 4.0 complete turns at a cons

Question

A record player rotates a record at 45 revolutions per minute. When the record player is switched off, it makes 4.0 complete turns at a constant angular acceleration before coming to rest. What was the magnitude of the angular acceleration (in rads/s2) of the record as it slowed down

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Ngọc Khuê 6 months 2021-08-13T09:09:43+00:00 1 Answers 2 views 0

Answers ( )

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    2021-08-13T09:11:27+00:00

    Answer:

    The  angular acceleration is \alpha =  0.4418 \ rad /s^2

    Explanation:

    From the question we are told that

          The  angular speed is w_f  =  45 \  rev / minutes =  \frac{45 *  2 *  \pi  }{60 }= 4.713 \  rad/s

           The  angular displacement is  \theta =4  \ rev  =  4 * 2 *  \pi =  25.14 \ rad

    From the first equation of motion we can define the movement of the record as

          w_f ^2  =  w_o ^2  +  2 *  \alpha *  \theta

    Given that the record started from rest w_o  =  0

    So

           4.713^2  =  2 *  \alpha *  25.14

            \alpha =  0.4418 \ rad /s^2

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