Show that the flywheel inside the train counteracts lean in a curve

In summary, the trolley should be directed in the negative x direction to counteract the rolling of the carriage.
  • #1
Karl Karlsson
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Summary: Consider a train carriage rolling along a curve that forms a left turn on the track. The carriage speed is directed along the y-axis (into the plane of the paper) in the figure. The trolley will have a tendency to curl in the curve in the specified direction. A flywheel is inserted which rotates with a large angular velocity ω0 around its axis of symmetry along the x-direction. Decide whether ω0 should be directed in the positive or negative x direction to counteract the rolling of the carriage.

Skärmavbild 2019-10-03 kl. 18.47.02.png

Consider a train carriage rolling along a curve that forms a left turn on the track. The carriage speed is directed along the y-axis (into the plane of the paper) in the figure. The trolley will have a tendency to curl in the curve in the specified direction. A flywheel is inserted which rotates with a large angular velocity ω0 around its axis of symmetry along the x-direction. Decide whether ω0 should be directed in the positive or negative x direction to counteract the rolling of the carriage. Motivation is required.

Solution:

Skärmavbild 2019-10-03 kl. 18.47.36.png


I understand the whole solution up to the last line. My reasoning why the vector w0 should be directed in the negative x direction:
w1 is a constant. In order to
Skärmavbild 2019-10-03 kl. 18.47.36.png
something in the right must have the opposite direction compared to what it had before, this can also be seen from the cross product on line 5 in the solution. Since w1 is constant, it must be the angular momentum of the mass centra that has the opposite direction, when we calculated MG, the momentum of motion was directed along the x-axis, now it must be directed along the minus x-axis and hence the vector w0 must be directed in the negative x direction. Can anyone mathematically show why I'm wrong. Thanks!
 
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  • #2
I'm having some trouble understanding the description of your reasoning.

As the carriage makes a left turn, the flywheel changes its orientation. The vector ## \mathbf M_G## is the torque ("moment of force") that must be applied to the flywheel in order to cause the flywheel to change its orientation. This torque comes from the supports at each end of the axis of the flywheel. These supports are attached to the walls of the carriage. So, ## \mathbf M_G## is the torque that the carriage applies to the flywheel. Therefore, the flywheel applies a reaction torque of the same amount but in the opposite direction to the carriage. Thus, the flywheel exerts a torque ##- \mathbf M_G## to the carriage. You need to show that ##- \mathbf M_G## is in the direction that helps to prevent the carriage from tipping over if ##\boldsymbol \omega_0## is in the positive x direction (as shown in the figure).

Is this how you are thinking about it? If you understand the solution up to the last line, then you understand that ## \mathbf M_G## is in the positive y-direction if ##\boldsymbol \omega_0## is in the positive x-direction. (The solution is based on the assumption that ##\boldsymbol \omega_0## is in the positive x-direction. See the third line of the solution and also the figure.) Therefore, ##- \mathbf M_G## is in the negative y-direction. Is this the correct direction in order to help prevent the carriage from tipping?
 
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  • #3
TSny said:
I'm having some trouble understanding the description of your reasoning.

As the carriage makes a left turn, the flywheel changes its orientation. The vector ## \mathbf M_G## is the torque ("moment of force") that must be applied to the flywheel in order to cause the flywheel to change its orientation. This torque comes from the supports at each end of the axis of the flywheel. These supports are attached to the walls of the carriage. So, ## \mathbf M_G## is the torque that the carriage applies to the flywheel. Therefore, the flywheel applies a reaction torque of the same amount but in the opposite direction to the carriage. Thus, the flywheel exerts a torque ##- \mathbf M_G## to the carriage. You need to show that ##- \mathbf M_G## is in the direction that helps to prevent the carriage from tipping over if ##\boldsymbol \omega_0## is in the positive x direction (as shown in the figure).

Is this how you are thinking about it? If you understand the solution up to the last line, then you understand that ## \mathbf M_G## is in the positive y-direction if ##\boldsymbol \omega_0## is in the positive x-direction. (The solution is based on the assumption that ##\boldsymbol \omega_0## is in the positive x-direction. See the third line of the solution and also the figure.) Therefore, ##- \mathbf M_G## is in the negative y-direction. Is this the correct direction in order to help prevent the carriage from tipping?

You seem to understand exactly my reasoning. Yes, I understand that ## \mathbf M_G## is in the positive y-direction if ##\boldsymbol \omega_0## is in the positive x-direction. The moment of force from the flywheel on the carriage must then be ##- \mathbf M_G## directed in the negative y - direction, but then ##\boldsymbol \omega_0## must be in the negative x - direction in order to prevent the carriage to tip over. Right?
 
  • #4
Karl Karlsson said:
You seem to understand exactly my reasoning. Yes, I understand that ## \mathbf M_G## is in the positive y-direction if ##\boldsymbol \omega_0## is in the positive x-direction. The moment of force from the flywheel on the carriage must then be ##- \mathbf M_G## directed in the negative y - direction, but then ##\boldsymbol \omega_0## must be in the negative x - direction in order to prevent the carriage to tip over. Right?
I don't understand the last part where you say, "but then ##\boldsymbol \omega_0## must be in the negative x - direction in order to prevent the carriage to tip over."

Does everything in the following two figures make sense to you?

1570287469711.png


You can see that in order for the flywheel to apply a torque in the correct direction as shown in the figure on the right, ##\boldsymbol \omega_0## must be in the positive x-direction.
 
  • #5
Be sure that you understand how Newton's 3rd law comes into play. If the carriage exerts a force on the flywheel at one end of the axis of the flywheel, then the flywheel exerts an equal but opposite force on the carriage at that point. It's a good idea to draw the forces acting at the ends of the flywheel.
 
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Related to Show that the flywheel inside the train counteracts lean in a curve

1. How does the flywheel inside the train counteract lean in a curve?

The flywheel inside the train acts as a stabilizing force, keeping the train from leaning too much in a curve. This is due to the principle of angular momentum, where the rotating flywheel creates an opposing force to the centrifugal force that causes the train to lean.

2. Why is it important for trains to counteract lean in curves?

If a train were to lean too much in a curve, it could potentially derail, causing accidents and delays. By using a flywheel to counteract lean, trains are able to maintain stability and ensure safe travel for passengers.

3. How does the flywheel's mass and speed affect its ability to counteract lean?

The mass and speed of the flywheel are directly related to its ability to counteract lean. A heavier and faster spinning flywheel will have a greater angular momentum, resulting in a stronger stabilizing force.

4. Is the flywheel the only mechanism used to counteract lean in trains?

While the flywheel is a common method, there are other mechanisms that can be used to counteract lean in trains. Some trains use hydraulic systems or active suspension systems to maintain stability in curves.

5. Can the flywheel inside the train malfunction and cause the train to lean in curves?

While it is possible for the flywheel to malfunction, trains have multiple safety mechanisms in place to prevent this from happening. Additionally, regular maintenance and inspections ensure that the flywheel is functioning properly and able to effectively counteract lean in curves.

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