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PhysicsRotationTranslational + RotationalHard2 minPYQ_2023
PhysicsHardnumerical

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).

 

Question diagram: A uniform disc of mass 0 . 5 kg and radius r is projected wi
Answer:
54.00
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

Stream:JEESubject:PhysicsTopic:RotationSubtopic:Translational + Rotational
2mℹ️ Source: PYQ_2023

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