Introduction.

Some sports like gymnastics, diving, and ice skating have deliberate rotational kinematics built into the performance of their routines. An athlete leaps with a launch velocity and a rotational motion which can be altered so that the athlete rotates more quickly, then the rotation is slowed down for the landing. Rotational kinematics are governed by the laws of rotational dynamics just as translational kinematics (e.g.of a parabolic orbit) are governed by translational dynamics.

In other sports, rotational motions are just part of the background for doing other things. Our bodies are built for rotation around our joints and are not capable of direct translational motion. In order to move we must use rotational kinematics and dynamics.

You may have noticed that a person trying to maintain balance in a precarious situation will automatically begin to windmill her arms. This can happen at the end of a sprint race, or a gymnast may do so when she feels that she is about to fall off the balance beam.

Why aren't translational Kinematics and Dynamics [Newton's Laws] enough?

The kinematics and dynamics of a point mass successfully describe the major movements in sports. But people and sports equipment are NOT point masses; they have extended sizes and shapes. How does this influence motion? Of course, we could describe a baseball bat as composed of a multitude of point masses and consider all the forces acting on each point mass. This would include the influence of neighboring point masses. We can describe one point mass very simply. In collision theory, we have described two point masses interacting very well. However, describing the interaction of an infinite multitude of point masses is generally very difficult. If the points have a definite and fixed geometric relation, we can use the simple physics of rotation added to the translational kinematics and dynamics. The laws here are not new laws. We are still using Newton's Laws of dynamics but in a form that takes the integrated point masses into account.

We have already seen one example of this. The center of mass of the body in translational motion, moves as if all the mass were concentrated at that point. The CM of a body moves in a way dependent only on the launch velocity. Any other motions of rotation or even of explosion do not modify the CM motion. The long jumper may continue running or hang after launch. This does not alter the CM motion and won't make the CM go further.

Rotational Kinematics.

A point mass fixed by a string rotates around a fixed point. It's handy to describe the spatial position by radius and angle rather than x and y. In this case, r is costant but the angle q changes. The time rate of change of q is the angular velocity, w.

angular velocity, w = Dq/Dt.

the point mass has an instanteous velocity tangential to the circle that it is travelling. This velocity is related to the angular velocity by the relation:

v = r*w

The larger the radius, for a given w, the larger the velocity.

Angular acceleration, a,is the time rate of change of w.

a = Dw/Dt.

From these definitions, we can work out the formulas for angular velocities and angular displacements in a way completely parallel to the formulae for linear acceleration, velocity and displacement, e.g.

For constant a, w = at; q = 1/2 at², w² = 2aq.

For the equations to be valid, the angles must be expressed in radians. There are 2p radians = 6.28 in one complete turn around a circle [360 degrees].

Torque and how to get it.

To cause angular acceleration, we must apply a rotational type of force. This is called torque, t. If a force is applied directly to the center of rotation, no rotational motion is produced. The force must be applied off center. For a given force, the larger the lever arm, the larger the torque. To get the rusted wheel nuts off a car, the mechanic usually uses a wrench with a long handle so that he can apply a large torque. The torque varies as the product of the force and the lever arm.

In the torque produced by a force on a long handled wrench, only the force perpendicular to the handle produces torque. The force along the axis of the handle either compresses or stretches the wrench but produces NO torque.

t = F * length where the length is perpendicular to the force.

Several torques can be applied to a body. Only the net torque which is the vector sum of the torques produces angular acceleration. For rotations about the elbow, the biceps produces torque in one direction and the triceps produces torque in the opposite direction. If the torques are balanced, no angular acceleration will occur.

Rotational Dynamics.

The dynamical law governing changes in angular speed is:

t = I a which is completely analogous to Newton’s 2nd Law: F = mA

Both of these are vector equations. If you want to change the angular momentum, you must apply a torque. Your muscles apply a force through a radius supplying the rotational torque to the joints. The larger the force, the bigger the torque and the time rate of change of angular momentum. The new concept is that the translational inertia, m has been replaced by I, the moment of inertia. Unlike, m, an athlete can modify I. This can be done to enhance athletic performance. If you want to run fast, you want your legs to to have high angular. For a fixed torque, this can be accomplished by decreasing I. [Decreasing the moment of inertia, I, CAN be done even though we cannot decrease m.] To decrease the rotational inertia, a runner will flex the leg at the knee joint to effectively shorten the leg as it is swung through. The lifting of the heels to almost touch the buttocks is universally practiced for sprinters.

What is I?

For a point mass, m, pivoting about a point at a radius, r, the moment of inertia is:

I = mr².

If you can change the effective radius of the mass by a factor of 2, there will be a 4X effect on the moment of inertia.

• Note that for a given torque supplied by a muscle, the quickness of the rotation depends on the smallness of the moment of inertia.
• Dt = I Dw/t . For a given rotation, and torque, the time to carry out the action is directly dependent on I. Decrease I by x2 and the time will be shortened by 2X.

An athlete has control over this. For a quick turnover of the legs for sprinting, the legs should be flexed to reduce the moment of inertia.

For translational motion, the inertia is proportional to the mass ONLY.

• For rotational motion, the inertia for rotation is proportional to the mass as well as the distribution of the mass.
• carrying a 30 lb mass which is concentrated like a lead brick is much easier than carrying and maneuvering a 30 lb racing shell which is distributed over 27’.
• A diver changes her rotational velocity greatly as her posture changes from a pike to an extended position.
• A sprinter has much quicker feet when the knee joint is fully flexed so that the foot is almost touching the buttock during the swing through of the leg.]

Rotational inertia in other sports.

Heavy shoes are anaethema to quick motion. Even though the mass of shoes is not large, their location at a large radius from the hips gives them a rather large moment of inertia. Track and field athletes use very very light track shoes. Similarly, while the bike wheels are not very heavy the rim and tires are at a large radius and so contribute heavily to the inertia of a bike. To be quick, the light rims and tires [sew-ons] are desired.

A use of rotational dynamics to estimate strategy for the throwing motion.

Q. Since a longer radius produces a larger tangential velocity for a given w, why is that a pitcher will throw the ball with his arm bent at the elbow?

A. The pitcher can generate a certain amount of torque with his shoulders over a fixed angle. The smaller the moment of inertia, the larger a and w can be. This argues for a short radius. Let's ignore the moment of inertia of the arms for the moment and make the calculation using only the mass of the ball. For the arm bent at the elbow, the ball is at radius r so I = mr².

w² = 2aq = 2 t/I q = 2 t/(mr²) q.

Since the tangential velocity v = rw, we have that v² = r²w² = 2tq/m which is independent of the radius. You cannot get a higher speed by extending the arm in this situation. However, higher speeds can be obtained by developing the muscle force to provide larger t as well as having a larger angular range of motion to obtain large q. For a defensive throw like a shortstop throwing to first base, the shortened arm motion will be quicker which gives a better chance to throw the runner out. Since there is no loss in ball speed for this motion, the shortened arm motion is desired. Soccer players with quick kicks should also employ this strategy.

Conservation of angular momentum

L = Iw

where w is the rotational velocity and I is the moment of inertia.

In the absence of external torques, a system of masses must have a constant angular momentum. The axis for the angular momentum is perpendicular to the plane of the rotation.

A skater obtains a rotational speed with her arms and legs held away from the vertical axis of rotation. As she draws her limbs inward, the mass doesn't change but the moment of inertia gets smaller. The result is that the angular velocity increases. This happens so that the angular momentum will be conserved.

Calculations.

L = I w where w is the angular velocity = Dq/Dt, and I is the moment of inertia. The moment of inertia of a point mass rotating at a distance r from a fixed point is I = mr².

e.g.a skater spinning at 90 revolutions per minute has an angular velocity of 90 rpm. To carry out calcuations in international units, we must convert this to radians/sec. 1 revolution = 6.4 radians and 1 minute = 60 sec.

w = 90 rpm * 6.4/rev * min/60 sec = 9.6 radians/sec

Energy in Rotation

If a wheel is spinning but its CM is stationary it still has kinetic energy. Setting a wheel spinning requires energy (work).

The kinetic energy of rotation is:

K = 1/2 Iw²

Rotational energy can be converted into gravitational energy. A rolling wheel can climb a hill. A gymnast performing the floor excercise can jump to an amazing height. This comes from storing energy in two ways. The gymnast gains by the trampoline effect because the floor is somewhat elastic. The gymnast also builds up quite a lot of rotational energy before converting this energy into height.

A comparison of the dynamical and conserved quantities in linear and rotational motion:

More general considerations of angular momentum.

Does the angular momentum of the wheels provide stability to a bike?

To understand this, we must consider a more general case of angular momentum.

L = Iw

and I = mr² and w = v/r where v is the tangential velocity.

So L = mr²*v/r = mv*r = p*r where v is the linear velocity and p is the linear momentum. Any body with linear momentum can be considered to have an angular momentum relative to some fixed axis; r is the perpendicular distance to the p vector where the p vector goes through the cm.

While the bike wheels have angular momentum, this is small compared with the L of the bike rider, who has a large mass and a large momentum.

eg. L_rider = (70 kg)*(10 m/s)*(1 m) = 700 kg-m²/s

The L_wheel = (1 kg * 0.5²) * 10/0.5 = 5 kg-m²/s, which is 100x less than the angular momentum of the rider. Clearly the dominant angular momentum is that of the rider, not the wheels.

Reverse Steering.

A rider going fast wants to make a precise and quick turn to the right. Which way should the handle bars be turned? It turns out that the handlebars should be flicked to the left to initiate the turn. Riders of bikes and motorbikes know this instinctively even if they don't do it explicitly.

Why does it work this way? When you are going straight and you flick the handle bars to the right, you will lean to the left. The leaning to the left causes a torque that will initiate a left hand turn! Remember the L of the rider is very large and so a large torque is necessary to change directions.

Effects of spin on the translational motion of balls

The jump shot in basketball is made with a deliberate backspin. Because of the rather low velocities of the launches, the backspin makes little difference to the flight of the ball in the air. However, should the basketball hit the rim or backboard, the backspin will tend to make the bounce more vertical by reducing the horizontal velocity of the ball. This action increases the likelihood that the ball will enter the hoop for a score.

How does this work?

The ball can experience frictional forces that act parallel to the surface it contacts. Consider a ball dropped vertically with clockwise spin. Rather than bouncing straight up, the ball will bounce off to the right and up. When it hits the floor the transverse velocity of the ball's surface will push the floor leftward. The reaction of the floor on the ball will be to push the ball to the right. If there were no friction, the ball would bounce straight up, spin or no.

A frictionless ball launched leftward with V_ox would continue with V_ox after the bounce. A ball with friction would suffer force toward the right and come off the bounce with reduced V_ox. If the ball were also to be launched with backspin [CW rotation] it would suffer an even greater reaction force from the floor and have its V_ox further reduced. This is just the effect desired for the backspin jump shot in basketball.

With topspin, the V_ox is increased upon touchdown to the ground. In basketball, this is used when applying English to the ball for layups from under the basket. In this case the shooter cannot get position to angle the ball properly to enter the basket, but the spin and the resultant reaction of the backboard launches the ball into the basket.

The spin in tennis.

In tennis, most strokes will strike the ground before being played. This provides a lot of opportunity for the use of spin to alter the bounce. A serve or groundstroke with topspin has the advantage of curving downward thereby giving a good chance of being played fair. At the same time the increased downward velocity makes the ball bound higher. The offensive player by altering the spin can make the flight and bounce of the ball as unpredictable as possible to enhance the chance of winning the point. An even more vicious effect is that the topspin will make the ball shoot forward with bigger horizontal velocity after the bounce. This effect is particularly strong with courts having high resistance like clay or hard courts. On grass courts, the grass has low resistance. This means that V_x is less affected by the bounce. For this reason, the grass courts are considered fast. Of course, the grass does not add to the V_x, but the V_x is bigger than on courts with high resistance. Winning tactics vary on each kind of court surface and it is extremely difficult to be a master at all of the tactics.

Producing spin on a struck ball.

Whenever a ball collides obliquely with a surface that has friction, the ball's parallel velocity is altered and a torque is applied to the ball by the off-center force. A tennis racket with its grid of strings on a compliant ball has a lot of friction and lots of potential for introducing spin. The upward and forward swing for a ground stroke launches a ball with topspin. You can think of the racket as brushing the ball upwards. Another way of thinking about it is to place yourself as an observer at rest on the racket head. This observer will see the ball approaching with a downward velocity. The ball-racket collision thereby produces a upward reaction force on the ball's surface or a CCW torque thereby producing a topspin. The downward chopping motion stroke produces backspin which is desirable for the softly hit ball which will just rise over the net and fall and then bounce with little or no V_x to make it unhittable. Sideways motions of the racket as used in the serve will give the ball both top and sidespin [like the 3/4 arm curve ball] to further confound the opponent.

Questions

1. If a mass on a string of radius r is subject to an angular acceleration, a for a period of t, what is the angular and tangential velocity of the mass at this time. Let r = 0.5 m, and a = 4 /sec².



2. What is the necessary angular acceleration for your foot to achieve a tangential speed of v when pivoting about your hip joint with a constant angular acceleration over an angle q? The foot and hip are separated by a distance r. Let v = 10 m/s andq = 1.2 radians



3. If an athlete reduces the moment of inertia of his leg by bending his knee to shorten the leg to 2/3 of the original leg length, how much faster can the athlete rotate his leg through Dq?



4. To have a quick release of the throw to 2nd base to catch a steal, how does the catcher flex his arm?



5. Estimate the moment of inertia of a diver in a pike position vs an extended position. How does this affect the speed of rotation. What position should be used for the entry to the water for a high scoring dive?



6. A wheel with mass = 5 kg, I = 30 kg m² is spun with w = 200 radians/sec. It is then permitted to run up a hill with friction. How high up the hill can this wheel go? Does the slope of the hill matter?



7. A sprinter after the finish line sometimes will swing arms in a circle to keep his balance. Which way should the arms be swung? Which way would you tend to fall if you didn't swing the arms?