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Abbandon
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1. A uniform rigid rod of length L and mass M is attached to a frictionless pivot point at one end. The rod is initially held completely horizontal and is released from rest. What is the linear and angular velocities at the instant the rod is in the vertical position?2. -ΔU = ΔK
3. -ΔU = ΔK(trans) + ΔK(rotat)
MgΔh = .5MV^2 + .5Iω^2 and I = (1/3)ML^2 since it is a rod pivoted at the end
MgΔh = .5MV^2 + .5((1/3)ML^2)ω^2 and Lω = V -------> V = (ω/L)
2gΔh = V^2 + (1/3)V^2
2gΔh = (4/3)V^2
V = ((3/2)gΔh) ^ (1/2) and ω = V/L = (((3/2)gΔh)^(1/2))/L
Is this correct? It then says to confirm your answer by using two photogates at the vertical position. The info for that problem is:
Length of the rod: 1m
Time it takes to get to the vertical position: 0.4419s
Time it takes to pass between the photogates: 0.0032s
Distance between the photogates: 0.016m
Mass of the rod (doesn't matter): 92.4g
3. -ΔU = ΔK(trans) + ΔK(rotat)
MgΔh = .5MV^2 + .5Iω^2 and I = (1/3)ML^2 since it is a rod pivoted at the end
MgΔh = .5MV^2 + .5((1/3)ML^2)ω^2 and Lω = V -------> V = (ω/L)
2gΔh = V^2 + (1/3)V^2
2gΔh = (4/3)V^2
V = ((3/2)gΔh) ^ (1/2) and ω = V/L = (((3/2)gΔh)^(1/2))/L
Is this correct? It then says to confirm your answer by using two photogates at the vertical position. The info for that problem is:
Length of the rod: 1m
Time it takes to get to the vertical position: 0.4419s
Time it takes to pass between the photogates: 0.0032s
Distance between the photogates: 0.016m
Mass of the rod (doesn't matter): 92.4g