Question

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

Answer 1

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`