\(\begin{array}{l} a = (\mu \;g \;cos \Theta ) + (g \;sin \Theta) \\\Rightarrow 0.5 * 10 * 0.8 + 10 * 0.6 = 10m/s^{2} \\\Rightarrow \alpha = \frac{(\mu\; mg \;cos\Theta) R}{\frac{1}{2}\;mR^{2}} = \frac{2\mu\; g\; cos\Theta}{R} \\\Rightarrow \frac{2 * 0.5 * 10 * 0.8}{0.4} = 20 rad/s^{2} \end{array} \)
Pure rolling will start when, v = RΩ or at = R (Ω 0 – at)
Therefore, 10 t = 0.4 (54 – 20 t)
t = 1.2 seconds.
The time taken by the cylinder to start pure rolling is 1.2 seconds.
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