JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Rotational Mechanics

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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).

Answer: 54

Solution

While the disc slides, kinetic friction decelerates its centre at

a = μg = 0.3 \times 10 = 3

^{- 2}

, so after

2

s the centre moves at

v = 18 - 3 \times 2 = 12

^{- 1}

, and at this instant the motion is pure rolling. The kinetic energy of a rolling disc is

\frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}

with

I = \frac{1}{2} m r^{2}

and

ω = \frac{v}{r}

, which reduces to

\frac{3}{4} m v^{2}

. Substituting

m = 0.5

kg and

v = 12

^{- 1}

, the total kinetic energy is

\frac{3}{4} \times 0.5 \times 144 = 54

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Integer answer

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JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Rotational Mechanics

Solution

More JEE Main questions on Rotational Mechanics