A flywheel revolving at 200 revolutions per minute is slowed down uniformly at the rate of 2 revolutions/min². The number of revolutions/s made by it before coming to a stop is

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Here is the solution of your question-

First, we arrive at the initial angular velocity by the statement of 200 rev/min = {200*2π}/60. Acceleration is negative. By applying First equation or rotational motion, and substituting the values, we get time required to stop the flywheel which will be π/3(answer of first part).

Then to find out the time period(T) of one revolution, put initial angular velocity as 2π/T(general formula). Then finally to find out the number of revolutions in this time= given time/time period of one rev. The answer will be 10π/9 (answer of second part).

Hope this helps you.

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