Acceleration of the disc, a=−μkg=−3ms−2.
Now, velocity of the centre of mass
v=u+at=18−3×2
⇒v=12ms−1
Now total kinetic energy during pure rolling(v=ωr) will become,
KE=21mv2+21Iω2=21mv2+212mr2r2v2
⇒KE=43mv2
=3×18=54J
JEE Main 2023 — Physics Mechanics
A uniform disc of mass 0.5kg and radius r is projected with velocity 18ms−1 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=10ms−2).

Held on 30 Jan 2023 · Verified 6 Jul 2026.
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