Send the link below via email or IMCopy
Present to your audienceStart remote presentation
- Invited audience members will follow you as you navigate and present
- People invited to a presentation do not need a Prezi account
- This link expires 10 minutes after you close the presentation
- A maximum of 30 users can follow your presentation
- Learn more about this feature in our knowledge base article
Physics in Gymnastics
Transcript of Physics in Gymnastics
Gymnasts use physics in every skill they perform and on every event. For this project, we are looking specifically at the physics behind floor skills like a standing back tuck and a running full twist. This prezi outlines several physics concepts in gymnastics like sample kinematics, forces, work and angular momentum.
GYMNAST'S WEIGHT: 50 or 60 kg
GYMNAST'S HEIGHT: 1.67 m
Full Twist Video
Centripetal force is present during the gymnast's rotation 1 meter above the ground. Assume the gymnast jumps straight up, tucks her body in, spins, and lands straight on her feet again. She rotates with her knees pulled into her chest forming a circle with a diameter of 1 meter.
Standing Back Tuck
As a gymnast performs a standing back tuck, she jumps straight up, rotates backward, tucks in her leags, and finishes rotating until her feet land on the ground again.
Let's assume the girl is 1.67 meters tall and jumps off the ground until her head reaches its maximum height of 2 meters. The girl starts her jump from rest and immediately accelerates at 0.4 m/s2 as soon as she leaves the ground.
So how can we calculate the gymnast's final velocity once she reaches her maximum height?
we can use:
since we know:
If the gymnast jumps 1 meter off the ground:
PHYSICS RWA: Gymnastics
By: Alina Pituch, Alexa Pituch, and Stacee Mendonca
Friction is a force that resists movement between two surfaces. It acts in the opposite direction of the way the object wants to slide. It is proportional to the force that presses the surfaces together, as well as the roughness of the surfaces. Therefore, to determine the force of friction, you need to know the resistance between the two surfaces, or mu, and the normal force perpendicular to the surfaces.
F(friction) = μN
If the gymnast weighs 6okg and the coefficient of friction between her feet and the spring floor is 0.6 then we can calculate the force of friction between her feet and the floor:
F(friction) = (0.6)(60kg)(9.8m/s2)
Angular momentum is the amount of rotation an object has and depends on the object's mass, shape, and speed. The closer the distribution of mass is to the axis of rotation, the faster the object rotates.
When a gymnast does a full twist, she pulls her arms close to her body so she spins faster and can complete the twist. By finding the distribution of her mass inc omparison with her axis of rotation, along with how fast she is spinning, we can calculate her angular momentum.
Calculating Angular Momentum in a Full Twist
In order to calculate angular momentum, we need to know how fast the gymnast rotates. If it takes her about 1 second to complete the full twist, then we know her rotational velocity. We also need to know her moment of inertia, which is defined with respect to her axis of rotation and her mass. For this example, we will estimate her moment of inertia as that of a cylinder. To find the angular momentum, we multiply the rotational velocity by the moment of inertia.
Free Body Diagram:
There are several forces present in both a standing back tuck and a full twist. As you can see in the standing tuck full body diagram, there are four main forces acting on the gymnast:
downward force of gravity, constantly acting on the gymnast
upward normal force acting on the gymnast when her feet are touching the ground
rightward force of friction preventing the gymnast's feet from sliding off the ground as she jumps up
upward force of the gymnast's jump
Linear momentum is present during the running passes in gymnastics. In the example of the running full twist pass, the gymnast builds momentum in her run.
p= mass x velocity
If the gymnast runs with a velocity of 2 m/s and has a mass of 50 kg, her linear momentum just before she begins her pass is:
p= (50 kg)(2 m/s) = 100 kg*m/s
LINK TO ONLINE QUIZ
L = I*(omega)
where L is the angular momentum, I is the moment of inertia, and omega is the rotational velocity
L = 1/2MR2 * 5.2m/s
L = 1/2(60kg)(1.67/2m)(5.2m/s)
L = 130 m2/s