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Question:

A cubical block of side 30 cm is moving with velocity 2 m/s 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:

13.3

5.0

9.4

96.7

Solution:

Angular momentum of the block about the point O will be constant.
Hence, mvr = Iω
Moment of inertia about the point O = ma²/6 + m(a√2)²
ω = mvr/I
After putting values:
ω = m × 2 × 0.15 / (2/3 × m × a²) → ω = 5 rad/s