Figure Skating and Physics #3. Angular Momentum

3. Angular Momentum

Figure skaters spend a lot of time rotating through the air or spinning on the ice. They score more when they can spin or rotate faster and longer.  To spin faster and longer, the skater must develop a large amount of angular momentum. Angular momentum is the amount that a body is rotating about a point throughout the jump. Angular momentum is generated by the skater applying a force against the ice, then the ice applies a ground reaction force on the skater and this ground reaction force gives the angular momentum. 

Angular momentum is represented by the equation L=IW, where I equals the moment of rotational inertia about the spin axis and is equal to 0.5mr^2 where m is mass and r is radius. W represents angular velocity, which means how fast the skater is spinning. The moment of inertia depends on the mass of an object and also the distribution of the mass around the axis of rotation. 

                                       

The conservation of angular momentum is the principle that the angular momentum of an object remains constant as long as there is no external torque or moment acts on the object. The equation for the conservation of angular momentum is I1W1 (initial) = I2W2 (final).  For instance, the moment of inertia decreases, the angular rotation has to increase to keep the same angular momentum. This is evident when a figure skater spins. A skater starts the spin with arms stretched (large moment of inertia). As the skater brings the arms in (decreasing the moment of inertia), the rotational speed increases. This is how those skaters like Mao Asada and Yuna Kim can perform incredible spins and jumps, along with many years of practice. 

Question ) What is the angular momentum of a figure skater spinning (with arms close to her body) at 4.4 rev/s, assuming her to be a uniform cylinder with a height of 1.6m, radius of 16cm and mass of 42 kg? ( Answer in kgm^2/s)

Figure Skating and Physics #2 - Projectile motion and Newton's Third Law

2. Projectile Motion and Newton’s Third Law

Newton’s third law of motion states that for every action, there is an equal and opposite reaction. Skilled figure skaters propel themselves across the ice in controlled motions by taking advantage of Newton’s third law of motion.

At the start of the movement, when the skater pushes off against the ice, the ice pushes back. This is how the skaters are able to perform their T-stops at the end by using friction to push in the opposite direction of their motion.

This principal also applies to jumps. Figure skaters body becomes projectiles when they jump moving both vertically and horizontally. When a skater forcefully pushes down, the ice forcefully pushes up, launching the skater into the air to perform the jumps. 

Moving across the ice, The skater’s horizontal velocity stays constant throughout the move compare to the vertical velocity (the skater goes up to a certain height and gravity pulls the skater back down) and you can notice that the forces are independent of each other. 

The height and distance of a figure skating jumps are determined by a skater’s take-off velocity, take-off angle, and take-off position. A skater generates take off velocity by applying force against the ice. These two formulas allows to calculate jump height and distance from take-off velocity. (Vf = Vi + at), (d= Vit + 1/2 at^2) 

It is very important to separate the object’s vertical take-off velocity from their horizontal take-off velocity because the gravity immediately starts slowing the skater’s vertical velocity, when the skater leaves the ice, but gravity has absolutely no effect on the skater’s horizontal motion since it is acting only in the vertical direction. When the skater is at the top of the jump, the vertical velocity is always zero. 

The vertical and horizontal velocity can be calculated by applying these concepts:

If you are given : 

-   V to (take-off velocity) 

θ (angle) 

Then you can get :

-   Vv (vertical velocity) 

Vh (horizontal velocity) 

By using sine and cosine :

Vv = V to sinθ

Vh = V to cosθ

These formulas come from the right triangle. (SOHCAHTOA)

Vertical Velocity— Vertical velocity is an indication of how quickly an object is falling or rising. Gravity immediately starts slowing the skater’s vertical velocity when the skater leaves the ice. Once gravity has slowed the skater’s upward velocity to zero, gravity then accelerates the skater back to the earth.

Question ) If Hannah’s take-off velocity is  10m/s, and the take-off angle is 45’, then what is Hannah’s vertical velocity?

Horizontal Velocity — Horizontal velocity is the speed of something flat on the ground like a car on the road. In order to jump straight up, the skater need to have some amount of vertical velocity and to move horizontally the skater need some amount of horizontal velocity.

Question) If Sue’s take-off velocity is 11m/s, and the take-off angle is 45’ then what is Sue’s horizontal velocity?

+ Question for the take-off velocity ) If Sarah’s horizontal velocity is 6 m/s, the vertical velocity is 6m/s, and the take-off angle is 45’, then what is Sarah’s take-off velocity?

Vertical Displacement 

Vertical displacement is the height of the jump. 

Take-off velocity is needed in order to calculate the vertical displacement. The easiest way to calculate is to first calculate the time it took to get from take-off to the top of the jump. Time up can be calculated from the equation Vf= Vi+at. Once the time-up have calculated, then we can get the vertical displacement from another equation D = Vit + 1/2 at^2.  

Question) Ashley is performing lutz jump at the Nationals. Her take off velocity is 12m/s and the take-off angle is 45’. What’s the height of the lutz jump?

Horizontal Displacement

Horizontal displacement is the distance between the skater’s take-off point and the landing point.

Horizontal displacement is calculated similarly to vertical displacement. You need to know the time for the jump from take-off to landing to calculate the horizontal displacement. 

In figure skating, the skaters take-off and land on a level surface, however, I’m gonna measure the distance from the centre of mass, not the surface of the ice. Thus, only if the centre of mass of the skater is at the same height at take-off and landing will the time up and time down be equal. The centre of mass depends on the position of the skater. The equation D = Vit + 1/2 at^2 will be used to calculate the Horizontal Displacement. 

Question) Gracie is performing flip jump at the ISU competition. If Gracie’s take-off velocity is 12m/s, take-off angle is 43’, and the time-up is 0.69s. Calculate the horizontal displacement for Gracie’s entire jump. 

Figure Skating and Physics #1 Friction

  Figure skating is a mixture of art and athleticism. No other sport combines the elegance with the danger of complex triple jumps, spiral sequences, lifts, and throws at thrilling speeds. Figure skating is a perfect example that you can apply to physics in everyday life. Here’s a perfect chance to watch examples of basic scientific concepts such as friction, angular momentum and projectile motion. 

1. Friction

The difference between dancing on a floor and skating on ice is the lack of friction. Compared to a wooden floor, ice has much less friction. Smooth ice provides very little resistance against objects, like ice skates, being dragged across its surface. Ice skates are designed to take advantage of this with their grooved blades that are only one-eighth of an inch thick, because of the minimal contact with an already low friction surface, skaters find little resistance as they slide against the ice and can glide smoothly. Friction is a force that opposes the sliding of two surfaces across one another. Friction arises because the molecules on both surfaces bond with each other, and resist when the surfaces try to move away and break the bonds. The more rough and jagged something is, the more easily more of its molecules will come into contact with molecules on the surface it touches, and thus the greater force of friction will exert. The general low level of friction on ice allows a skater to glide along the surface smoothly without friction stopping the motion as soon as it’s begun. In order to reduce friction between the ice and the skater, it is important to keep the blades sharpened. 

Issac Newton’s first law of motion, which explains an object in motion tends to stay in motion unless acted on by a force. This is the reason why ice skaters whose motion isn’t being acted on by a powerful enough force of friction, tend to stay in motion, unless they use force to stop themselves. Skating would be impossible if there were no friction at all on the ice. It is the friction between the skate and the ice when a skater pushes off that starts the motion to begin with. Friction also allows a skater to stop. 

The coefficient of friction U  is the ratio of the force of friction(Ff) to the normal force(Fn).The equation is Uf = Ff/Fn. Coefficient of kinetic friction(Uk) is the ratio of kinetic friction to the normal force. The corresponding equation is Uk=Fk/Fn. We will only need to use this equation (Uk=Fk/Fn) to solve the question below. 

                                         

Question) Meryl glides across the ice on her skates, which have steel blades. She is 40kg. What is the magnitude of the force of friction acting on Meryl? (Uk = 0.01 - steel on ice)