Q82 / 90 · JEE Main 2023 (30 Jan Shift 2) · Physics

A uniform disc of mass 0.5 kg and radius r is projected with velocity 18 m s-1 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 s-2).

Figure 

Correct answer: 54

Solution

Acceleration of the disc, a=-μkg=-3 m s-2.

Now, velocity of the centre of mass

v=u+at=18-3×2

⇒v=12 m s-1

Now total kinetic energy during pure rolling(v=ωr) will become,

KE=12mv2+12Iω2=12mv2+12mr22v2r2

⇒KE=34mv2

=3×18=54 J

Hence, the correct answer is 54.

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