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-

Answer 1

Answer:1.6 rad/s

Explanation:

Given

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

radius of disc $r=0.25 m$

Force $F=8 N$

Torque $T=I\alpha =F\cdot r$

$5\times \alpha =8\times 0.25$

$\alpha =0.4 rad/s^2$

using

$\theta =\omega _0\times t+\frac{\alpha t^2}{2}$

$\pi =0+\frac{0.4t^2}{2}$

$2\pi =0.4t^2$

$t^2=5\pi$

$t=\sqrt{5\pi }$

$t=3.96 s$

$\omega =\omega _0+\alpha t$

$\omega =0+0.4\times 3.96$

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