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A disk with a rotational inertia of 5.0 kg. m2 and a radius of 0.25 m rotates on a frictionless fixed axis perpendicu-

lar to the disk and through its center. A force of 8.0 N is applied tangentially to the rim. If the disk starts at rest,
then after it has turned through half a revolution its angular velocity is :
(1) 0.57 rad/s
(2) 0.64 rad/s
(3) 1.6 rad/s
(4) 3.2 rad/s

User Varantir
by
8.3k points

1 Answer

7 votes

Answer:1.6 rad/s

Step-by-step explanation:

Given

moment of Inertia of disk
I=5 kg-m^2

radius of disc
r=0.25 m

Force
F=8 N

Torque
T=I\alpha =F\cdot r


5* \alpha =8* 0.25


\alpha =0.4 rad/s^2

using


\theta =\omega _0* t+(\alpha t^2)/(2)


\pi =0+(0.4t^2)/(2)


2\pi =0.4t^2


t^2=5\pi


t=√(5\pi )


t=3.96 s


\omega =\omega _0+\alpha t


\omega =0+0.4* 3.96


\omega =1.58 rad/s\approx 1.6 rad/s

User Abdan Syakuro
by
8.3k points