Answer
a). $0.0331rad/s^{2}$
b). $1.8\times 10^{-3} m/s^{2}$
Work Step by Step
a). Angle rotated by the ball in travelling 2.5 m = $\frac{2.5}{3\times 10^{-2}}=\frac{250}{3} rad$
Now, $\omega^{2}=\omega_{0}^{2}-2a\theta$
So, $0=2.35^{2}-2a\frac{250}{3}$
or, $a=0.0331rad/s^{2}$
b). $v^{2}=u^{2}-2aS$
$0^{2}=(0.03\times2.35)^{2}-2a(3.2)$
or, $a=7.766\times 10^{-4}m/s^{2}$
Thus, net tangential acceleration = $a+r \times angular\,\, acceleration $
=$7.766\times 10^{-4}+ (0.03\times 0.0331)=1.8\times 10^{-3} m/s^{2}$
