A solid cylinder having radius \(R\) and length \(L\) is slipping on a rough horizontal plane. At time \(t=0\) the cylinder has a translational velocity \(v_0=49\) m/s, perpendicular to its axis and a rotational velocity \(v_0/4R\) about the centre. The time taken by the cylinder to start rolling is ________ seconds. (coefficient of kinetic friction \(\mu_K=0.25\) and \(g=9.8\) m/s\(^2\))

Initial translational velocity, \(v_0 = 49\) m/s.

Initial angular velocity, \(\omega_0 = \dfrac{v_0}{4R}\).

Velocity of the lowest point of the cylinder is \(v_{bottom} = v_0 - \omega_0 R = v_0 - \dfrac{v_0}{4} = \dfrac{3v_0}{4} > 0\).

Since the lowest point is slipping forward, kinetic friction \(f_k = \mu_k m g\) acts in the backward direction.

Translational acceleration, \(a = -\dfrac{f_k}{m} = -\mu_k g\).

Angular acceleration, \(\alpha = \dfrac{f_k R}{I} = \dfrac{\mu_k m g R}{\dfrac{1}{2} m R^2} = \dfrac{2\mu_k g}{R}\).

At time \(t\), the translational velocity is \(v = v_0 - \mu_k g t\).

The angular velocity is \(\omega = \omega_0 + \alpha t = \dfrac{v_0}{4R} + \dfrac{2\mu_k g}{R} t\).

For pure rolling to start, \(v = \omega R\).

\(v_0 - \mu_k g t = \left( \dfrac{v_0}{4R} + \dfrac{2\mu_k g}{R} t \right) R\)

\(v_0 - \mu_k g t = \dfrac{v_0}{4} + 2\mu_k g t\)

\(3\mu_k g t = \dfrac{3v_0}{4}\)

\(t = \dfrac{v_0}{4\mu_k g}\)

Substituting the given values \(v_0 = 49\) m/s, \(\mu_k = 0.25\), and \(g = 9.8\) m/s\(^2\):

\(t = \dfrac{49}{4 \times 0.25 \times 9.8} = \dfrac{49}{9.8} = 5\) s.

Answer: \(5\)