Last updated on October 25, 2020
This chapter gives you an overview of all the basic physics that are important in biomechanics. The aim is to give you an introduction to the physics of biomechanics and maybe to awaken your fascination for physics. The topics mass, momentum, force and torque are covered. In order to illustrate the physical laws in a practical way, there is one example from physics and one from biomechanics for each topic area. Have fun!
1.Mass
In physics, mass M is a property of a body. Its unit is kilogram [kg] .
Ein Rennauto hat z.B. eine Masse von M = 800kg .
The volleyball player has a mass of M = 60kg
2. Momentum
In physics, the momentum p describes the state of motion of a body. Its unit is kilogram times meter per second [\frac{kg * m}{s}] The momentum combines the mass with the velocity. The momentum is described by the following formula:
p = M * vHere p is the momentum, M the mass and v the velocity of the body.
Sometimes the formula sign for the impulse is also written like this: \vec{p} . This is because the momentum is a vectorial quantity and therefore has a size and a direction. For the sake of simplicity we write only p .
If you shoot a light and a heavy bullet at the same speed against the door of our racing car, the heavy bullet will leave the bigger and deeper dent. This is because the impulse of the heavy ball is greater.
3. conservation of momentum
Let us assume that a billiard ball rolls over a billiard table at a constant speed and then hits a second ball that is lighter than the first ball. After the collision, the first ball will stop and the second ball will continue rolling, but the speed of the second ball is greater than that of the first. This happens because of the principle of conservation of momentum, which says that the momentum before the collision must be equal to the momentum after the collision. Let us assume that the velocity of the first ball is v_1 = 2 \frac{m}{s} and the ball has the mass M_1 = 0.2kg and the mass M_2 of the second ball is only half as large at 0.1kg. Then it follows from the conservation of momentum that the second (lighter) ball has a velocity v_2 twice as high as the first ball. The calculation path would be as follows:
p_1 = p_2
M_1 * v_1 = M_2 * v_2
v_2 = M_1 ∗ \frac{v_1}{M_2} = \frac{P_1}{M_2}
v_2 = \frac{0.4}{0.1} [\frac{kg∗\frac{m}{s}}{kg}] = 4 \frac{m}{s}
So, the velocity of the second ball is v_2 = 4 \frac{m}{s} .
The principle of conservation of momentum also applies to the beach volleyball player's attack. The impulse of the stroke hand and arm is transferred to the ball. The impulse p_s of the stroke hand and arm is calculated from the hand and arm weight M_H = 4kg and the hand arm speed v_H = 2 \frac{m}{s} . p_s is therefore 4kg * 2 \frac{m}{s} = 8 \frac{kg * \frac{m}{s}}{kg} . The speed of the ball v_B after the stroke is calculated from \frac{p_s}{M_B} , where M_B is the mass of the ball of 0.2kg , and is therefore v_B = 40 \frac{m}{s} . So, the volleyball is faster after the stroke than the arm was moved during the stroke movement because of its smaller mass.
4. Force
In physics, the force F describes an action that causes a body to be accelerated or deformed. Its unit is Newton N . The force combines mass with acceleration. The formula for the force is
F = M * aHere F is the force acting on a body, M is the mass of the body and a is the acceleration the body experiences due to the force F .
Also forces are vectorial quantities with a direction, which is why they are sometimes written like this: \vec{F} .
For the acceleration of a = 2 \frac{m}{s^2} from our race car on the first 100 meters, the engine had to generate a force. This force can be calculated from the mass of the car M = 800kg and the acceleration a = 2 \frac{m}{s^2} .
F = M * a
F = 800kg * 2 \frac{m}{s^2}
F = 1600 [kg ∗ \frac{m}{s^2}] = 1600N
So the car engine generated a force of F = 1600N to accelerate the racing car with the mass of M = 800kg by a = 2 \frac{m}{s^2} . With a light car less power would have been needed to achieve the same acceleration.
For the jump at the attack stroke, the beach volleyball player has to generate power with her leg muscles to accelerate her mass (body weight) upwards. Since the mass of the beach volleyball player is constant, the more force she can generate the greater the acceleration a of her mass and the higher she will jump. This also means that lighter players can jump higher if they are able to generate the same jumping force with their legs.
4. Torque
In physics, torque D describes the rotational effect of a force on a body. Its unit is Newtonmeter [Nm] .
The formula for the torque is
D = F ∗ s = M ∗ a ∗ sHere D is the torque, F the force, s the distance between the force and the center of rotation, M the mass and a the acceleration.
Let us remember our childhood and the moments we spent on seesaws in the playground. When two children with the same mass sit on the seesaw, the seesaw will be in balance and both children will be at the small height. If one child now slides further backwards on his side of the seesaw, the seesaw on this side will tilt down. What happens here is that the child that slides back will increase its distance to the axis of rotation (the middle of the seesaw) and thus extend its lever. As a result, its torque increases and exceeds that of the other child. The seesaw therefore starts to rotate.
Mathematically the situation can be described as follows: Both children have the same mass M and both are subject to the same acceleration a = 9.81 \frac{m}{s^2} . Consequently, the two forces generated by the children are equal to F_1 = F_2 . So the torques of the children depend only on the distance s to the centre of rotation, which here is the distance of the child to the centre of the seesaw. Because one child slides backwards its distance becomes greater s_1 > s_2 , which means that its torque is greater D_1 > D_2 .
In our musculoskeletal system, the muscle generates the force and transfers it to the bone with its tendon. So the muscle pulls on the bone. Depending on where the muscle is attached to the bone, the distance s to the joint axis varies. This also has a strong influence on the torque that the muscle generates. During the attack stroke, the muscles of the upper body produce a force that pulls on the humerus via a tendon. This rotates the club arm around the shoulder and accelerates from back to front where it finally hits the ball.
