A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s^-2 • Its net acceleration in ms^-2 at the end of 2.0s is approximately

Given: Radius of disc, R = 50 cm angular acceleration $\alpha =2.0{\mathrm{rads}}^{-2};\text{ time }\mathrm{t}=2\mathrm{s}$.

angular acceleration $\alpha$ = 2.0 rad s^-2; time t = 2s Particle at periphery (assume) will have both radial (one) and tangential acceleration 

$a_t=R\alpha =0.5\times 2=1\mathrm{m}/{\mathrm{s}}^2$

From equation

$\begin{array}{l}\omega ={\omega }_0+\alpha \mathrm{t}\\ \omega =0+2\times 2=4\mathrm{rad}/\sec \\ {\mathrm{a}}_{\mathrm{c}}={\omega }^2\mathrm{R}=(4{)}^2\times 0.5=16\times 0.5=8\mathrm{m}/{\mathrm{s}}^2\end{array}$

Net acceleration, 

$a_{\text{net }}=\sqrt{a_t^2+a_c^2}=\sqrt{1^2+8^2}\approx 8\mathrm{m}/{\mathrm{s}}^2$