Torque and angular momentum of a swing

In summary: I am thinking that h could be 1/3ML^2 and L could be the distance from the gymnasts centre of gravity to the bar.
  • #1
dimpledur
194
0

Homework Statement


A gymnas is performing a giant swing on the high bar. IN a simplified model of the giant swing, assume that the gymnast keeps his arms and body straight. Also assume that the gymnast does no work. With what angular speed should he be moving at, at the bottom of the swing, to ensure that he can make it all the way around? The distance from the bar to his feet is 2m and his centre of gravity is 1m in.



The Attempt at a Solution



I'm not sure if I am supposed to apply conservation of energy for this question, or if there is a simpler way?

I was going to say:

Fnet=mv^2/r
Fnet*r/m = v^2
sqrt (Fnet*r/m) = v

and then convert v into angular velocity.

However, I don't know how to solve for Fnet etc.

I'm not sure how to approach this problem at all. Any help is greatly appreciated.
 
Physics news on Phys.org
  • #2
Actually, I may have figured it out. Correct me if I am wrong, please.

Since the question states that no work is done by the man swinging, then there is no translational kinetic energy. However, there is rotational kinetic energy.

Therefore, we can state:

Rotational kinetic energy at the bottom = potential energy at the top
Krot = Ut
1/2Iw^2 = mgh
w2 = 2mgh/I

We can have the man swinging assume the shape of a rod, where I = 1/3ML^2

so
w^2 = (2mgh)/(1/3ML^2)
w^2 = 6gh/L^2
w = sqrt (6gh) / L

now I am not sure if L is the distance to the centre of gravity, or if it is the entire length of the rod. However, I assumed it was the distance of the entire rod.

so, w = sqrt (6*9.8*4) / 2 = 7.67rad /s


?
 
  • #3
perhaps the question is unclear?
 
  • #4
Well, it seems we are in the very same class and are having the very same problem. I came up with the same solution as yourself, and am having a similar problem completing the question.

I am unsure of what values to use for h and L
 

FAQ: Torque and angular momentum of a swing

What is torque?

Torque is a measure of the force that causes an object to rotate around an axis. It is calculated by multiplying the force applied to an object by the distance from the axis to the point where the force is applied.

How does torque affect a swing?

Torque is what allows a swing to move and gain momentum. When a person pushes or pulls on a swing, they are creating a torque that causes the swing to rotate around its axis. The greater the torque, the faster the swing will move.

What is the difference between linear and angular momentum?

Linear momentum is the measure of an object's motion in a straight line, while angular momentum is the measure of an object's rotational motion. In the case of a swing, linear momentum would refer to the back and forth motion of the swing, while angular momentum would refer to the rotational motion around the swing's axis.

How is angular momentum conserved in a swing?

According to the law of conservation of angular momentum, the total angular momentum of a system remains constant as long as there are no external torques acting on it. In the case of a swing, the angular momentum is conserved as long as there is no external force causing it to stop or change direction.

What factors affect the torque and angular momentum of a swing?

The torque and angular momentum of a swing can be affected by several factors, including the force applied to the swing, the distance from the axis where the force is applied, the mass of the swing, and any external forces such as friction or air resistance.

Back
Top