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QJEEM16Ph28e6: JEE Main 2016 Physics - Class 11 JEEM16Ph28e6 Solution: A cubical block of side 30 cm is moving with velocity 2 ms −1…

A cubical block of side 30 cm is moving with velocity 2 ms−1 on a smooth horizontal surface. The surface has a bump at a point O as shown in figure. The angular velocity (in rad/s) of the block immediately after it hits the bump, is :

JEE Main 2016 (Online) 9th April Morning Slot Physics - Rotational Motion Question 180 English

Multiple Choice Options

  • A 5
  • B 6.7
  • C 9.4
  • D 13.3

Step-by-Step Verified Solution

Before hitting point 0,

angular moment = mv ×\times a2{a \over 2}

After hitting point 0,

Angular momentum = Iω{\rm I}\omega

∴\therefore     Iω{\rm I}\omega = mva2{{mva} \over 2}

⇒\Rightarrow     ω\omega = mva2I{{mva} \over {2{\rm I}}}

I{\rm I} = moment of inertia about edge,

=    ma26+m(a2)2{{m{a^2}} \over 6} + m{\left( {{a \over {\sqrt 2 }}} \right)^2}

=    ma26+ma22{{m{a^2}} \over 6} + {{m{a^2}} \over 2}

=    2ma23{{2m{a^2}} \over 3}

∴\therefore     ω\omega = mva2×2ma23{{mva} \over {2 \times {{2m{a^2}} \over 3}}} = 3v4a{{3v} \over {4a}} = 3×24×0.3{{3 \times 2} \over {4 \times 0.3}} = 5 rad/s

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