GATE 2022 ME (ME1) – Question 51
A cylindrical disc of mass $m = 1$ kg and radius $r = 0.15$ m was spinning at $\omega = 5$ rad/s when it was placed on a flat horizontal surface and released (refer to the figure). Gravity $g$ acts vertically downwards as shown in the figure. The coefficient of friction between the disc and the surface is finite and positive. Disregarding any other dissipation except that due to friction between the disc and the surface, the horizontal velocity of the center of the disc, when it starts rolling without slipping, will be _________ m/s (round off to 2 decimal places).
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Correct answer: 0.24 to 0.26
Explanation
The friction force acts at the contact point, so the angular momentum about the contact point is conserved: $\frac{1}{2}mr^2\omega = \frac{1}{2}mr^2\omega' + mr^2\omega'$, which gives $\omega' = \frac{\omega}{3}$. Rolling without slipping then has $v = r\omega' = \frac{0.15 \times 5}{3} = 0.25$ m/s.