Rotational Kinematics: Falling Rod

AI Thread Summary
A uniform rigid rod of length L and mass M, pivoted at one end, is released from a horizontal position, and the discussion focuses on determining its linear and angular velocities when it reaches the vertical position. The initial potential energy is converted into kinetic energy, but there is a debate about the correct interpretation of the height change (Δh) and the linear velocity (V) at the vertical position. One participant clarifies that Δh should be L, as only the end of the rod moves, while the pivot remains stationary, impacting the potential energy calculation. The discussion emphasizes the importance of considering the center of mass and acknowledges that the rod's motion is purely rotational about the pivot. The conversation highlights the need for precise definitions and understanding of the system's dynamics to derive the correct equations for velocity.
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
 
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Abbandon said:
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?

I do not think your derivation is correct. First, you did not specified what Δh and V are.
If you consider the motion of the rod with respect to the pivot then it is rotation about a fixed axis. KE=Iω2, I = 1/3 ML2.

ehild
 
ehild said:
I do not think your derivation is correct. First, you did not specified what Δh and V are.
If you consider the motion of the rod with respect to the pivot then it is rotation about a fixed axis. KE=Iω2, I = 1/3 ML2.

ehild

Sorry for not being clearer:
Δh = L since h(i) = 0 and h(f) = L
and V is the linear velocity at the vertical position
 
Last edited:
The change of the potential energy of the uniform rigid rod is not mgL. Think: the piece at the pivot does not change height, while the height of the opposite end changes by L.

Linear velocity of what piece of the rod do you speak about ?

ehild
 
HINT: 1.consider center of mass OF THE ROD ?
{--- BODY HAS NO TRANS. KINETIC ENERGY}
 
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