Answer:[tex]\alpha =9.302\ rad/s^2[/tex]
Explanation:
Given
mass of sphere [tex]m=250\ gm[/tex]
diameter of sphere [tex]d=4.30\ cm[/tex]
radius [tex]r=\frac{4.30}{2}\ cm[/tex]
[tex]f=0.0200\ N[/tex]
friction will provide resisting torque so
[tex]f\times r=I\times \alpha [/tex]
where [tex]I=\text{moment of Inertia}[/tex]
[tex]f=\text{friction force}[/tex]
[tex]\alpha =\text{angular acceleration}[/tex]
[tex]I=\frac{2}{5}mr^2[/tex]
[tex]0.02\times r=\frac{2}{5}mr^2\times \alpha [/tex]
[tex]\alpha =\frac{5}{2r}\times f[/tex]
[tex]\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02[/tex]
[tex]\alpha =9.302\ rad/s^2[/tex]
(b)time taken to decrease its rotational speed by [tex]21\ rad/s[/tex]
[tex]t=\dfrac{\Delta \omega }{\alpha }[/tex]
[tex]t=\dfrac{21}{9.302}[/tex]
[tex]t=2.25\ s[/tex]
