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QJEEM23Ph2be4: JEE Main 2023 Physics - Class 11 JEEM23Ph2be4 Solution: A uniform disc of mass and radius is projected with velocity at …

A uniform disc of mass 0.5 kg0.5 \mathrm{~kg} and radius rr is projected with velocity 18 m/s18 \mathrm{~m} / \mathrm{s} at t=0\mathrm{t}=0 s on a rough horizontal surface. It starts off with a purely sliding motion at t=0 s\mathrm{t}=0 \mathrm{~s} . After 2 s2 \mathrm{~s} it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after 2 s2 \mathrm{~s} will be __________ J\mathrm{J} (given, coefficient of friction is 0.30.3 and g=10 m/s2g=10 \mathrm{~m} / \mathrm{s}^{2} ).

JEE Main 2023 (Online) 30th January Evening Shift Physics - Rotational Motion Question 54 English

Step-by-Step Verified Solution

v=v0−μgtv = {v_0} - \mu gt

⇒v=18−0.3×10×2=12\Rightarrow v = 18 - 0.3 \times 10 \times 2 = 12 m/s

⇒\Rightarrow Kinetic energy =12mv2+12mv22= {1 \over 2}m{v^2} + {1 \over 2}{{m{v^2}} \over 2}

=34mv2=34×0.5×144 J=54 J= {3 \over 4}m{v^2} = {3 \over 4} \times 0.5 \times 144\,\mathrm{J} = 54\,\mathrm{J}

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