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

Answer: 54. Friction deceleration = μ g = 3 m/s², so after 2 s: v = 18 - 3 × 2 = 12 m/s. (Check: for a disc starting with pure sliding, v = (v₀)/(1+k) = 18/1.5 = 12 m/s.) KE of rolling disc = 3/4mv² = 3/4× 0.5 × 144 = 54 J.

A uniform disc of mass 0.5 kg and radius r is projected with velocity… | JEE Main 2023 (30 Jan, Shift 2)

Q47JEE Main 2023 · 30 JAN · SHIFT 2Difficult

Rolling Motion

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

Numerical answer type. Work it out, then open the answer.

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Answer: 54

Friction deceleration $= \mu g = 3\ \text{m/s}^2$ , so after 2 s: $v = 18 - 3 \times 2 = 12$ m/s.
(Check: for a disc starting with pure sliding, $v = \dfrac{v_0}{1+k} = \dfrac{18}{1.5} = 12$ m/s.)
KE of rolling disc $= \frac{3}{4}mv^2 = \frac{3}{4}\times 0.5 \times 144 = 54$ J.

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Rolling Motion question from JEE Main 2023 (30 Jan, Shift 2)

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