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JEE Main 2016 (9 Apr, Shift 1) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from JEE Main 2016 (9 Apr, Shift 1)

Q70JEE Main 2016 · 9 APR · SHIFT 1Difficult

Angular Momentum & its Conservation

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:

Figure

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Answer: (a)

CvOa/2a = 0.3 m
  1. The impulsive force acts at O, so angular momentum about O is conserved; the block then pivots about the edge at O.
  2. Before: L=Mva2L = Mv\frac{a}{2} (centre of mass at height a2\frac{a}{2}).
  3. After: IO=Ma26+M(a2)2=23Ma2I_O = \frac{Ma^2}{6} + M\left(\frac{a}{\sqrt{2}}\right)^2 = \frac{2}{3}Ma^2.
  4. ω=va/22a2/3=3v4a=3(2)4(0.3)=5 rad/s\omega = \frac{va/2}{2a^2/3} = \frac{3v}{4a} = \frac{3(2)}{4(0.3)} = 5\ \text{rad/s}
  5. So the answer is (a) 5.0.
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