Why is the elastic potential energy in position 2 zero?

In summary, at the first position, the potential energy is maximum because the torsional spring is uncoiled. However, at the second position, the potential energy is maximum because the torsional spring is coiled.
  • #1
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Homework Statement
The torsional spring at A has a stiffness of
k = 2000 N /rad and is uncoiled when theta = 0°. Determine
the angular velocity of the bars, AB and BC, when theta= 0°, if
they are released from rest at the closed position, theta = 90°.
The bars have a mass per unit length of 20 kg/m.
Relevant Equations
conservation of energy
1673095011819.png


Hello,

so we have two potitions right, if we take ##\theta = 90## as the first position (i.e. both rods are flat) and then the second position at ##\theta = 0##.
I totally understand the exercise, not difficult. The only issue I am having is the torsional spring... it says that it is uncoiled at 0 degrees. Does this mean that the potential energy at position 2 is at maximum?
Because if so I get a negative value from which I need to find ##\omega_{AB}## which is of cousre not possible.

this is the eq. by conservation of energy:
##T_1 + V_1 = T_2 + V_2## note: angular velocity of BC is zero ! that is why there is only one term.
##0 + 0 + = 450*\omega_{AB}^2 + \frac122000*(\pi/2)^2 + 2060.1##
##\omega_{AB}^2 = -...... ##
Which is not possible. And from this I know that the elastic potential energy needs to be at the other side, but the question is why is the EPOT maximum at position 1 i.e. theta = 90 degrees as it says that it is uncoiled at theta = 0 degrees?

Thanks in advance.!
 
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  • #2
Are you not worried about the change in gravitational potential energy of the center of mass of each bar as the configuration changes?
simphys said:
but the question is why is the EPOT maximum at position 1 i.e. theta = 90 degrees as it says that it is uncoiled at theta = 0 degrees?
It's the change in potential energy that counts. Regardless of where the zero of elastic potential energy is chosen, it increases when the end of the spring attached to the bar is rotated from the relaxed position.
 
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  • #3
simphys said:
it says that it is uncoiled at 0 degrees. Does this mean that the potential energy at position 2 is at maximum?
No, uncoiled (position 2) would be the minimum EPE.
 
  • #4
The angular velocity of the bars AB and BC when theta= 0° should be zero, as both reach their stroke’s limits.
Perhaps the question is about an instant before.

Like in the previous thread, the decreasing spring moment is fighting decreasing moments induced by the weights of the bars, as those move from a horizontal towards a vertical position.

The rotational inertia of the bars may need to be considered as well.
I believe that a pure energy approach can’t properly solve this problem, not sure.
 
  • #5
Lnewqban said:
The angular velocity of the bars AB and BC when theta= 0° should be zero, as both reach their stroke’s limits.
Perhaps the question is about an instant before.

If that little ledge weren't there the bars could move beyond ##\theta = 0##, and even with it there they could probably move slightly past ##\theta = 0## . Once they reach ##\theta= 0## the torque reverses direction.

Lnewqban said:
The rotational inertia of the bars may need to be considered as well.
I believe that a pure energy approach can’t properly solve this problem, not sure.
I definitely think the OP is to consider the rotational inertia of the bars. The angular velocity of link BC is determined as a function of the angular velocity of link AB. I think COE is fine...if applied with care (perhaps great care).
 
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  • #6
simphys said:
##0 + 0 + = 450*\omega_{AB}^2 + \frac122000*(\pi/2)^2 + 2060.1##
Hi @simphys. Have you sorted this out yet? If not, can I add to what has already been said.

The equation above is difficult to understand because you have used values rather than symbols. But (after trying it myself) it appears that:
##450*\omega_{AB}^2## is the gain in BC’s translational kinetic energy;
##\frac12 2000*(\pi/2)^2## is the loss of the spring’s EPE;
##2060.1## (J) is the gain in AB's and BC's gravitational potential energy.

If so, you have forgotten to include AB’s rotational kinetic energy.
(You have correctly taken BC’s rotational kinetic energy as zero though without clear explanation.)

Also, you have incorrectly formulated your equation and/or have a sign problem. The left side of your equation is zero but the right side is positive!
 
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  • #7
Steve4Physics said:
If so, you have forgotten to include AB’s rotational kinetic energy.
(You have correctly taken BC’s rotational kinetic energy as zero though without clear explanation.)

BC should have both rotational and translational kinetic energy?

EDIT: Never mind, not when ##\theta = 0^{\circ}##
 
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