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Torque in a non inertial reference frame 
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#1
Mar2512, 09:06 PM

P: 261

Hi,
This is partly inspired by the questions in the thread about normal forces on a cornering car but I thought I'd post here instead of mix that thread up with my question. So suppose we have a biker leaning into a corner. There are the normal force and the force of friction which act at the contact point between the wheel and the road. The weight of the biker acts downward from the CG. Let's consider the contact between the wheel and the road as our axis of rotation. If I look at it from the non inertial frame of the bike, there are two torques: The weight and the centrifugal force and they are in opposite directions so they can balance each other out and the biker's lean angle remains constant. However, if we look at this from an inertial frame, there is only one torque: The torque which arises due to the weight and this should make the biker lean even more (and eventually fall). There is nothing to compensate this torque so what am I missing here? How should I correctly analyze this from an inertial frame? Thank you! 


#2
Mar2512, 09:26 PM

P: 10

Because the cyclist's wheels are rotating about an axis perpendicular (or at least nonparallel) to the axis that defines the corner the cyclist is going around.; you're missing the fact that there is rotational motion present.
If the cyclist were to somehow manage this corner with stationary wheels, then they would indeed have problems staying upright... 


#3
Mar2512, 09:32 PM

P: 1,011

In both cases (inertial and noninertial), the vertical vectors should match. In the inertial case however, the difference between the horizontal friction force on the bike (by the ground) and the centripetal force on the bike is equal and opposite to the difference between the horizontal friction force on the ground (by the bike) and the reactive centrifugal force on the ground. In other words, in the inertial case, we have not a centrifugal force on the bike, but a centripetal one created by the ground acting on the bike toward the axis of the turn. You will have to consider 8 vectors and not just 4. 


#4
Mar2512, 09:36 PM

P: 1,011

Torque in a non inertial reference frame



#5
Mar2512, 10:03 PM

P: 1,011

By the way "gyroscopic stability" is simply the result of the fact that when you apply a force at a right angle to a body, the velocity of masses try to align with that force. The reason is obvious with force that is applied in the same direction for a long enough time; it will cause a change in velocity aligned with that force. By doing this, subsequent changes in displacement tend to align with that sustained force. This requires work to be done.
In the case for unconstant forces, if the force on mass points on the gyroscope was maintained at right angles to the motion of the mass points, the body or field which is acting on each mass point must be positioned such that the line between that body or field and each mass point is aligned with the forces between them, so as to obey Newton's laws of motion. This requires work to be done. In this case, it is not done onto the gyroscope. Instead it is done on the body acting on the gyroscope in order to obtain the proper position for generating forces on the mass points at a right angle. Therefore, the reason why gyroscopes (and other gyroscopelike devices) can be "gyrostabilized" is not some mystical property of precession and rotating reference frames. It has to do instead with the conservation of energy. A convincing reasoning which demonstrates this fact is the existence of counterrotating gyros, which do in fact have the ability to stabilize objects, despite the fact that their precession effects cancel. A notable example of a technology which has variants employing counterrotating gyros is called the gyro monorail. 


#6
Mar2512, 10:10 PM

P: 10

And I'm not being mystical, thanks, just brief! 


#7
Mar2512, 10:27 PM

P: 1,011




#8
Mar2512, 11:28 PM

P: 1,011

A skilled rider can easily bank on a turn with this contraption. The reason why it stays upright is because the motion along the turn provides a centripetal force on the bike by the ground and a centrifugal force of the bike due to the tendency of objects to maintain their trajectory. Together, these forces on the bike can provide the torque needed to keep bike from tipping over, provided it is sufficient. In a famous physics example: A wheel on a string behaves like this for three reasons: 1) In addition to the vertical tension on the string which counters the wheel's weight (though not the rotating force due to it), there is also a lateral tension on the bicycle's wheel by the string, establishing a centripetal force on the wheel, which generates a contribution to torque on the wheel's axis towards the "equator" around the string. 2) The spinning of the wheel enhances the centrifugal forces, providing an additional contribution to the lateral forces, by increasing the root mean square velocity of the bike mass along the circumference of the path taken by the wheel around the string. This also generates a contribution to torque on the wheel's axis towards the "equator" around the string. 3) The bicycle wheel must move around in a certain direction due to Coriolis forces acting on the wheel due to the interaction of the gravity with the momentum of the wheel. The stability of the wheel of the above experiment has nothing to do establishing the precession itself. All establishing the precession does is just change the direction (not magnitude) of the angular momentum at the wheel axle, which is due to the Coriolis effect. The reason why it doesn't cause the stability is this: The torque on the wheel due to gravity and the vertical tension of the string is capable of causing angular acceleration, if not countered. The "precessional torque" does not involve an angular acceleration but a change in spin axis orientation. The derivative of a constant magnitude angular momentum with respect to time (i.e. the torque) can be nonzero if the axis is precessing (changing orientation). However, such a torque does not cause the wheel to accelerate around the axis, nor can it do anything to resist the angular acceleration that may occur on the wheel. I repeat again: It is the extra lateral force which may provide the torque to counter gravity! 


#9
Mar2512, 11:33 PM

P: 261

Thank you for the replies but I still can't see it.
First the ice skates argument. As far as I can tell, it doesn't matter that my wheels are spinning. the.drizzle could you explain to me once again why the fact that wheels are turning matters here? Because if I look at the thing from the noninertial frame (i.e. the frame attached to the bike), there is a torque due to the centrifugal force [itex]mv^{2}/r*h[/itex] where h is the height of the CG. This is balanced by the torque due to the weight. Now, I didn't need to think about spinning wheels in this frame so I can't see why its relevant in an inertial frame. kmarinas, the reactive centrifugal force seems to be the force exerted by the bike on the road. This is also at the contact point between the wheels and the road and hence has no torque as our axis is at the same place. Another thing that I am confused about is that you are talking about two forces acting on the bike: A centripetal force and a horizontal friction force. But these are the same force. The friction is what gives the bike a centripetal force. As far as I can tell, there are only three forces on the bike: The weight, the normal force and the friction. Thank you both for your explanations, I do appreciate your help very much. 


#10
Mar2612, 12:28 AM

P: 1,011

The centrifugal force (due to the tendency of traveling in a straight line) exists across the entire wheel, but the centripetal force imposed by the ground only exists as a counterforce to the bottom side of the wheel. They therefore help each other in countering the torque due to gravity and the normal force. The bicycle is able to trace a corner because the centripetal force on the bike (synonymous with the friction force by the ground to the bike in this case) is stronger than the centrifugal force on the bike (reaction to changes in velocity). Therefore, a body in a turn can be both recipient of centrifugal and centripetal forces simultaneously. A common analogy you can witness on the road is called a rollover. The road provides a centripetal force due to friction, but the speed of the car overcomes this and the centrifugal force causes the vehicle to swerve outward, leading to a rollover with a direction of "rolling over" that depends on the torque caused by these forces. 


#11
Mar2612, 05:30 AM

P: 4,012

For simplicity you can consider a point mass on a massless beam attached via a hinge to a rotating beam.
In the rotating rest frame of the whole thing the torque from gravity around the hinge is constant, so if it wasn't canceled by the centrifugal torque, it would generate angular acceleration around that constant axis, and make the mass fall inwards. In the inertial frame however the torque from gravity is a vector which continuously changes its direction, and that makes matters complicated. Simplistically: after half a circle the torque vector from gravity is flipped, so it cancels all the angular acceleration it created on the opposite side of the circle. But in reality this transition happens continuously. This intuitive problem is similar to the question: Why doesn't the centripetal acceleration change the speed of an object in circular orbit, when it modifies the velocity vector? The reason is that the acceleration continuously changes direction to be perpendicular to velocity. But for torques it is ever more difficult to visualize. 


#12
Mar2612, 10:41 AM

P: 261

AT, that's a nice explanation. It took me a while but I can actually see what's going on now. Thank you very much!



#13
Mar2612, 12:01 PM

P: 1,011

For me, these sorts of explanations serve as mystifying simplicity. It's like saying that "God did it". I understand that it is easier to "explain" things that way, but really, I don't see how it actually explains anything. From what I understand, the basic explanation could actually be simpler than what you gave. All one has to do is look at what a car does when it is moving too fast on a turn. The spin of the top increases the tendency the travel in a straight line, and so it magnifies the ability to an continue outward path, via reactive centrifugal forces. This stabilizes the top because position and lateral acceleration of a spinning top are about 90 degrees out of phase as gravity tries to turn the top inwards. 


#14
Mar2612, 06:14 PM

Sci Advisor
P: 2,470

McLaren, if you go to inertial coordinate system, the pivot point and CoM actually accelerate. If you pick a static pivot, you'll notice that you do get torque, but the angular velocity of CoM also changes.
The simplest solution for this is to use CoM as your pivot. That's what you normally do when you deal with freebodies. In that case, acceleration of CoM is independent of angular motion of the body. Then looking at net torque produced by normal and centripetal forces, it is zero. And gravity can't produce torque because it acts on CoM. So that solves that. 


#15
Mar2612, 07:14 PM

P: 261

K^2, you're absolutely right. It is loads easier to consider the CoM or indeed switch to the frame of the bike itself. I was just curious why it seemed to be so hard in an inertial frame with the axis running through the contact point.
I see now that I failed to account for the fact that the circular motion of the bike gives us a change in angular momentum about my axis. I missed that term. And yet, the axis moves such that there is always a change in angular momentum about it but there is never any angular momentum. As A.T. said, that is analogous how there is a centripetal acceleration but never any centripetal velocity in uniform circular motion. That's quite cool now that I see it! Thank you everyone for your replies; it's such a basic mechanics problem but I feel like I learned something new! 


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