A uniform disc of mass 0.5 kg and radius r is projected with velocity 18 m/s at t = 0s on a rough horizontal surface. It starts off with a purely sliding motion at t = 0s. After 2s it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after 2s will be________J (given, coefficient of friction is 0.3 and g = 10 m/s²).

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answer is 54.

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Detailed Solution

using angular momentum conservation
$\begin{array}{l} 0.5 \times 18 \times r = 0.5 \times v \times r + \frac{{mr}^{2}}{2} ω \end{array}$

$\begin{array}{l} 18 = v + \frac{rω}{2}; v = rω 18 = \frac{3 v}{2} \Rightarrow v = 12 m / s \end{array}$

$∴ KE (at t = 2 \sec) = \frac{1}{2} \frac{(0.5) (12)^{2}}{2} \frac{r^{2}}{r^{2}}$

$= \frac{1}{2} [\frac{3}{2} (0.5) (12)^{2}] = 54 J$

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