Answer
The ball slides for a time of $~~1.2~s$
Work Step by Step
We can write an expression for the angular speed:
$\omega = \omega_0+\alpha~t$
$\omega = 0+\alpha~t$
$\omega = \alpha~t$
We can write an expression for the linear speed:
$v = v_0+at$
When the ball is rolling smoothly, $\omega = \frac{v}{R}$
We can use the expression for angular speed to write an expression for the linear speed:
$\omega = \alpha~t$
$\frac{v}{R} = \alpha~t$
$v = R~\alpha~t$
In part (b), we found that the linear acceleration is $a = -2.1~m/s^2$
In part (c), we found that the magnitude of angular acceleration is $\alpha = 47~rad/s^2$
We can equate the two expressions for $v$ to find $t$:
$v = R~\alpha~t = v_0+at$
$R~\alpha~t-at = v_0$
$t = \frac{v_0}{R~\alpha-a}$
$t = \frac{8.5~m/s}{(0.11~m)(47~rad/s^2)-(-2.1~m/s)}$
$t = 1.2~s$
The ball slides for a time of $~~1.2~s$
