Torque on rigid body when angular momentum is not constant

In summary, the conversation discusses the rotational motion of a rigid body around a fixed axis, specifically focusing on the case where the angular momentum vector is not parallel to the angular velocity vector. The example of a horizontal barbell rotating around a vertical axis is used to demonstrate the concept of uniform precession motion, where the component of angular momentum perpendicular to the axis follows a circular motion with the same angular velocity. The theorem of angular momentum is applied to show that there must be an external torque acting on the system, and the origin of this torque is debated, considering the weight force and the reaction of the barbell support. The relationship between the pivot point chosen for the calculation of angular momentum and the torque exerted by external forces is also discussed.
  • #1
Soren4
128
2
I 'd like to clarify some doubts about the rotational motion around a fixed axis of a rigid body, in the case the angular momentum vector [itex] \vec {L} [/itex] is not parallel to the angular velocity [itex] \vec {\omega} [/itex] .

In particular, consider a horizontal barbell with two equal masses [itex]m[/itex] , forced to rotate around a vertical axis [itex] z[/itex] not passing through the center of the barbell with a constant angular velocity [itex] \omega[/itex]. In this case, considering a point of the axis [itex]z[/itex] for the calculation of momenta (pivot point), [itex] \vec {L} [/itex] , which is the sum of the angular momenta of the two masses, performs a uniform precession motion. Therefore, while the component of [itex]\vec{L}[/itex] parallel to the [itex] z [/itex] axis [itex] L_ {z} [/itex] is constant, its component perpendicular to the axis [itex] L_ {n} [/itex] follows a circular motion with angular velocity [itex] \omega[/itex] .

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From the theorem of angular momentum, in this case, [itex] \vec {\tau} = \frac {d\vec {L}}{dt} \neq 0 [/itex], so there must be a torque [itex]\tau[/itex] generated by external forces. Since [itex] \vec {L}[/itex] follows a uniform precession motion [itex] \vec {\tau} = \frac {d \vec {L}}{dt} = \vec {\omega} \wedge \vec {L} \implies \tau = L_ {n} \omega [/itex] (1).

My main concern relates to the origin of the torque. The external force in question should be the weight force, which exerts a non zero torque, given the two different distance of the masses from the axis of rotation. However there is also the reaction of the barbell support, which should exert an equal and opposite torque, since the barbell remains in this position during rotation. Now if this is true, then the sum of these two torques is zero, but this can not be given that [itex] \vec {L}[/itex] is varying over time. I can not understand where I'm wrong: maybe I should not consider the reaction of the support?

In addition I would like to ask something about [itex] L_ {n} [/itex] too. This component of [itex] \vec {L} [/itex] in general depends on the pivot point selected for the calculation of the angular momentum, however (in the case of uniform precession) (1) holds. Does this mean that the torque applied by the external forces changes depending on how I choose the pivot for the calculation of [itex] \vec {L} [/itex] ? The torque [itex]\vec{\tau}[/itex] of the external forces calculated with (1) is to be interpreted as the torque with respect to the same pivot used for the calculation of [itex] \vec {L} [/itex] ?
 
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  • #2
Soren4 said:
My main concern relates to the origin of the torque. The external force in question should be the weight force, which exerts a non zero torque, given the two different distance of the masses from the axis of rotation. However there is also the reaction of the barbell support, which should exert an equal and opposite torque, since the barbell remains in this position during rotation. Now if this is true, then the sum of these two torques is zero, but this can not be given that ⃗LL→ \vec {L} is varying over time. I can not understand where I'm wrong: maybe I should not consider the reaction of the support?

i think i will advise you to see 'the motion of a top' analyzed in almost all textbooks to get a physical picture of precession and the necessary conditions for precession.
 

What is torque on a rigid body?

Torque on a rigid body is the measure of the force that causes an object to rotate around an axis. It is calculated by multiplying the force applied to the object by the distance from the axis of rotation.

How is torque related to angular momentum?

Torque and angular momentum are closely related, as torque causes a change in angular momentum. When the torque on a rigid body is not constant, the angular momentum of the object will also change over time.

What factors affect the torque on a rigid body?

The torque on a rigid body is affected by the magnitude and direction of the applied force, as well as the distance from the axis of rotation. The angle at which the force is applied also plays a role in determining the torque.

How is torque measured?

Torque is measured in units of Newton-meters (N*m) in the SI system. In other systems, it may be measured in units such as pound-feet (lb*ft) or pound-inches (lb*in). In order to measure torque accurately, the force and distance must be measured using appropriate instruments.

Can torque on a rigid body ever be negative?

Yes, torque can be negative. This occurs when the applied force and the direction of rotation are in opposite directions. In this case, the torque is considered to be in the opposite direction of the rotation. Negative torque can cause a decrease in angular momentum, resulting in a slowing down or reversal of the rotation of the object.

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