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$$^2$$)

Take rightward velocity as positive.

Given:
v₀ = 49 m/s
ω₀ = v₀ / (4R)

For pure rolling:
v = ωR

step 1: friction effects

kinetic friction:

f = μmg = 0.25 × mg

translational deceleration:

a = −f/m = −μg = −0.25 × 9.8 = −2.45 m/s²

step 2: angular acceleration

torque = fR

I (solid cylinder) = (1/2)mR²

$$\alpha=\frac{fR}{I}=\frac{μmgR}{(1/2)mR^2}=\frac{2μg}{R}$$

$$\alpha=\frac{2\times0.25\times9.8}{R}=\frac{4.9}{R}$$

step 3: write v and ω

$$v=49−2.45t$$

$$\omega=\frac{49}{4R}+\frac{4.9}{R}t$$

step 4: rolling condition

$$v=\omega R$$

$$49−2.45t=\frac{49}{4}+4.9t$$

step 5: solve

$$49−\frac{49}{4}=4.9t+2.45t$$

$$\frac{147}{4}=7.35t$$

$$t=\frac{147}{4\times7.35}=\frac{147}{29.4}=5$$