Answer:
[tex]a=0.0937 \ m/s^2[/tex]
Explanation:
Given that
Angular velocity (ω)= 0.6 rad/s
Angular acceleration (α)= 0.3 [tex]rad/s^2[/tex]
Radius (r)= 0.2 m
We know that is disc is rotating and having angular acceleration then it will have two acceleration .one is radial acceleration and other one is tangential acceleration.
So
[tex]Radial\ acceleration(a_r)=\omega ^2r\ m/s^2[/tex]
[tex]Radial\ acceleration(a_r)=0.6^2 \times0.2 \ m/s^2[/tex]
[tex](a_r)=0.072 \ m/s^2[/tex]
[tex]Tangential\ acceleration(a_t)=\alpha r\ m/s^2[/tex]
[tex]Tangential\ acceleration(a_t)=0.3\times 0.2\ m/s^2[/tex]
[tex](a_t)=0.06 \ m/s^2[/tex]
So the total acceleration ,a
[tex]a=\sqrt{a_t^2+a_r^2}\ m/s^2[/tex]
[tex]a=\sqrt{0.06^2+0.072^2}\ m/s^2[/tex]
[tex]a=0.0937 \ m/s^2[/tex]
