Question

Sol. [Key Concept: Friction gives linear deceleration $a = \mu_{k}g$ and angular acceleration $\alpha = \frac{2\mu_{k}g}{R}$; rolling starts when $v = \omega\ R$, solving to $t = 5$ s.]

$${a = \frac{\mu_{K}mg}{m} = \mu_{K}g }{\tau = I\alpha \Rightarrow \mu_{K}mgR = \frac{{MR}^{2}}{2} \times \alpha }{\Rightarrow \alpha = \frac{2\mu_{K}g}{R}}$$

$$\begin{array}{r} \therefore v = v_{0} - \mu_{K}gt\#(i) \end{array}$$

Also, $\omega = \omega_{0} + \frac{2\mu_{K}gt}{R}$

$$\frac{v}{R} = \frac{v_{0}}{4R} + \frac{2\mu_{k}gt}{R} $$Solving Eq. (i) & (ii)

$${\frac{v_{0}}{4} + 2\mu gt = v_{0} - \mu_{k}gt }{3\mu gt = v_{0} - \frac{v_{0}}{4} }{3\mu gt = \frac{3v_{0}}{4} \Rightarrow t = \frac{v_{0}}{4\mu_{K}g} = \frac{49}{4 \times \frac{1}{4} \times 9.8} = 5sec}$$.