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A wheel of moment of inertia 0.2 kgm 2 rotates at 360 rpm, about its vertical axis. What is the angular speed of the wheel, when a torque of -1Nm is applied about the same axis for 3s is
is your answer 3(4pi + 5) rad/sec?
TORQUE=MOMENT OF INERTIA*ALPHA(ANGULAR ACCELERATION) -1=0.2*ALPHA =-5m/s^2 appliying equation of motion for angle final omega=initial omega-alpha*time =120pi-0.5*3 =120pi-1.5
TORQUE=MOMENT OF INERTIA*ALPHA(ANGULAR ACCELERATION)
-1=0.2*ALPHA
=-5m/s^2
appliying equation of motion for angle
final omega=initial omega-alpha*time
=120pi-0.5*3
=120pi-1.5
12∏-15
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