Gymnastics max torque on the High Bar

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SUMMARY

The maximum torque experienced by a gymnast performing a giant on the High Bar occurs at approximately 90 degrees during the upward motion, primarily due to the gravitational force acting on the gymnast's center of mass (CoM). The torque is influenced by the gymnast's angular acceleration and the position of their hands on the bar. To optimize performance, gymnasts must manage the torque generated by gravity as they transition through the motion, adjusting their grip and body position to maintain control and speed.

PREREQUISITES
  • Understanding of torque and its vector equation
  • Knowledge of angular acceleration and its calculation
  • Familiarity with the concept of center of mass (CoM)
  • Basic principles of rotational dynamics in gymnastics
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  • Study the vector equation for torque and its application in gymnastics
  • Learn about angular acceleration and how to calculate it in rotational systems
  • Explore the role of center of mass in dynamic movements
  • Investigate techniques for optimizing grip and body positioning on the High Bar
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This discussion is beneficial for gymnastics coaches, sports scientists, and athletes seeking to enhance their understanding of torque dynamics and improve performance on the High Bar.

knapklara
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Homework Statement
Where is the maximum torque when a gymnast is doing a giant (circle around the bar)?
Relevant Equations
The giant goes counter clockwise. In my opinion the maximum torque is at 3/4 of the circle. How to further explain the answer?
The-backward-giant-circle-on-the-high-bar-The-gymnast-circles-from-the-handstand_Q640.jpg
 
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knapklara said:
Homework Statement:: Where is the maximum torque when a gymnast is doing a giant (circle around the bar)?
Relevant Equations:: The giant goes counter clockwise. In my opinion the maximum torque is at 3/4 of the circle. How to further explain the answer?

View attachment 301261

Are you familiar with the vector equation for torque? If not, what is the equation that you use to calculate torque, and what defines the angle in that equation? :smile:
 
It should be no torque on the bar, other than the induced by the friction between rotating hands and fixed bar, which gymnasts reduce as much as possible using powder and gloves.
Perhaps you refer to the pronounced bending effect of the rotating mass on the bar?
 
knapklara said:
Homework Statement:: Where is the maximum torque when a gymnast is doing a giant (circle around the bar)?
Relevant Equations:: The giant goes counter clockwise. In my opinion the maximum torque is at 3/4 of the circle. How to further explain the answer?

View attachment 301261
You are assuming a friction-free grip on the bar and a rigid man?

And you want the position where the force of the bar on the man's hand exerts the most torque about the man's center of mass?

It seems to me that there is a straightforward way to determine the torque supplied by the bar on the man's hands about the man's center of mass. But first one needs to know the man's angular acceleration.

There is a straightforward way to know the man's angular acceleration: Can you figure out the net torque on the man about the bar?
 
Of course, I haven't given enough detail. My question would be the biggest torque on the athlete. Do I make sense?
 
knapklara said:
Of course, I haven't given enough detail. My question would be the biggest torque on the athlete. Do I make sense?
About what reference axis? Unless the net force is zero, torque is only unambiguous if an axis is specified.
 
jbriggs444 said:
You are assuming a friction-free grip on the bar and a rigid man?

And you want the position where the force of the bar on the man's hand exerts the most torque about the man's center of mass?

It seems to me that there is a straightforward way to determine the torque supplied by the bar on the man's hands about the man's center of mass. But first one needs to know the man's angular acceleration.

There is a straightforward way to know the man's angular acceleration: Can you figure out the net torque on the man about the bar?
The gymnast in the accompanying diagram does not seem to be assumed rigid. It seems like they want us to consider how the CoM is changing in each position. I wonder if this is supposed to be qualitative exploration?
 
Last edited:
erobz said:
The gymnast in the accompanying diagram does not seem to be assumed rigid. It seems like they want us to consider how the CoM is changing in each position I wonder if this is supposed to be qualitative exploration?
That is a sensible interpretation. Maximum torque from gravity about the bar.
 
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jbriggs444 said:
That is a sensible interpretation. Maximum torque from gravity about the bar.
Indeed, it is this torque that provides angular acceleration, allowing the athlete to rotate faster around the bar. In order for this not to cancel out in between both sides, the athlete must change the torque by gravity when going up relative to when going down.
 
  • #10
Orodruin said:
the athlete must change the torque by gravity when going up relative to when going down.
Alternately, the athlete may arrange for the time interval going up to be less than the time interval going down. [The same sort of contortions that assist in the one tend to assist in the other, so it is difficult to separate out the two tactics]
 
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  • #11
I guess I'm meant to observe the athlete as a lever. I'm not sure how to give a better explanation since English is not my first language. Isn't it the biggest torque at about 90° angle if we consider gravity in the first half of the motion, kind of similar as when doing a biceps curl?
 
  • #12
knapklara said:
I guess I'm meant to observe the athlete as a lever. I'm not sure how to give a better explanation since English is not my first language. Isn't it the biggest torque at about 90° angle if we consider gravity in the first half of the motion, kind of similar as when doing a biceps curl?
I think you are to qualitatively compare the CoM of the gymnast (think about how his CoM is changing as he is extending and contracting), as well as the component of his weight perpendicular to the line joining his CoM to the axis of rotation for each position shown. If you think about it like this, the answer will become apparent.
 

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