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Help with a dynamics problem please?
Two cases are presented where a uniform square of side length, L, are allowed to fall from being balance on a corner, point O, to the edge. In case one, the block will rotate on a frictionless hinge. In case 2, the block will slide on a frictionless surface as it falls. The block has mass, m. The center of mass is positioned slightly to the right of being above O, so the blocks will fall in a clockwise direction.
For both cases:
1. With what velocity does the corner hit the ground?
2. What is the angular velocity of the block just before impact?
1 Answer
- SteveLv 71 decade agoFavorite Answer
The potential energy loss for each case is PE = m*g*y where y = (L/4)*(2 -√2)
Also, the moment of inertia of the square about its CM is m*L²/6
Case 1:
PE = KEt + KEr = ½m*V² + ½I*w², but w = V/r = 2V/(√2L). Combining,
(2/3)V² = g*(L/4)*(2 -√2), or V² = g*L*(3/8)*(2 -√2)→ V = .46868√(gL)
This is the velocity of the CM; the far corner will have twice that velocity.
Angular velocity will equal the corner velocity divided by L.
Case 2:
In this case the CM moves downward only. The final w = 2V/L. The rotational KE = ½Iw² = (1/3)mV².
KEtot = ½m*V² + (1/3)m*V² = (5/6)m*V²
PE = KEtot → m*g*(L/4)*(2 -√2) = (5/6)m*V² → V² = (3/5)((2 -√2)/2)*g*L → V = .41916√(gL)
Again, this is the velocity of the CM; the corner velocity Vc is twice that, and angular velocity is Vc/L
