13(3). A CERTAIN WHEEL TURNS THROUGH 90 REVOLUTIONS IN15 sec, ITS ANGULAR SPEED

Question

13(3). A CERTAIN WHEEL TURNS THROUGH 90 REVOLUTIONS IN15 sec, ITS ANGULAR SPEED AT THE END OF THE PERIOD BEING10 rev / sec. (A) WHAT IS THE ANGULAR SPEED OF THE WHEEL AT THEBEGINNING OF THE 15-sec INTERVAL, ASSUMING CONSTANT ANGULARACCELERATION ? (B) HOW MUCH TIME HAD ELASPED BETWEENTHE TIME THE WHEEL WAS AT REST AND THE BEGINNING OF THE 15-secINTERVAL ? REVISED PRINTING - FUNDAMENTALS OF PHYSICS ; .HALLIDAY ANDRESNICK ; ISBN 0471344311.... THE ANSWERS ARE (A) 2.0 rev/sec (B) 3.8 sec. 13(3). A CERTAIN WHEEL TURNS THROUGH 90 REVOLUTIONS IN15 sec, ITS ANGULAR SPEED AT THE END OF THE PERIOD BEING10 rev / sec. (A) WHAT IS THE ANGULAR SPEED OF THE WHEEL AT THEBEGINNING OF THE 15-sec INTERVAL, ASSUMING CONSTANT ANGULARACCELERATION ? (B) HOW MUCH TIME HAD ELASPED BETWEENTHE TIME THE WHEEL WAS AT REST AND THE BEGINNING OF THE 15-secINTERVAL ? REVISED PRINTING - FUNDAMENTALS OF PHYSICS ; .HALLIDAY ANDRESNICK ; ISBN 0471344311.... THE ANSWERS ARE (A) 2.0 rev/sec (B) 3.8 sec.

Explanation / Answer

angular displacement = 90 rev, time t = 15 s final angular velocity ' = 10 rev/s a) find initial angular velocity = ( + ')*t/2 = 2/t - ' = 2.0 rev/s angular acceleration = (' - )/t = 0.533rev/s2 b) find time t' initial " = 0 final ' = 2.0 rev/s = (' - ")/t' t' = (' - ")/ = 3.8 s

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