Hi again, I'm pushing this understanding of moment of inertia and the torque required to overcome it because it is important in biomechanics.
If you think about it every movement involves levers and pendulums associated with Rotational motion of a mass. The leg about the hip, the lower leg about the knee the foot about the ankle, the hallux about the 1st MPJ, the whole body Centre of Mass travelling over the planted foot. While several bodies might have similar mass, the distribution of that mass has a great influence on the effort required by the muscles to overcome the inertia of the mass in rotation.
We don't need to use maths in our daily practice but I think it will help to see the maths to get a more intuitive picture that we can use daily to estimate or visualize what's happening to our patient.
So we've seen that a mass has inertia that needs to be overcome by a force to induce motion. In linear problems the inertia can be found by the equation of Newtons 2nd law f=ma - the inertia is proportionate to the mass x acceleration and in the opposite direction to the force applied.
In the case of a mass rotating about an axis though, the inertia or the resistance to rotational motion is expressed as moment of inertia (I) and I=mr^2 and mr^2 is expressed in kg.m^2.
The torque (t) due to the moment of inertia is a function of the acceleration of the mass and so we get the equation t = Ia (where 'a' is the angular acceleration in radians/second - not linear acceleration in meters per second) NB a radian is equivalent to the length of the radius of interest. Since the circumference of a circle is Pi (3.14) x diameter, there are therefore 6.28 (approx) radians in 360dgs of the circumference of a circle and one radian = 57.3dgs..
In the earlier example of the sledge hammer, most of the mass is distributed at one end of the long handle and it is possible to ignore the mass of the handle itself if we are not being too picky. So, in this case the equation I=mr^2 works directly but in the case of shapes where the mass is more evenly distributed then the change in moment of inertia is not so large when considering the torque at various axes relative to the CoM. If you go here you can see that the mr2 or mL^2 (where L = length instead of radius) of the various shapes and their rotational axes is multiplied by a fraction.
So if the hammer head has a mass of 3kg and the handle is 500mm and the acceleration is 20rad/s^2 then 3x(0.5*0.5)x20 = 15Nm.
If we assume for convenience gravity is 10m/s^2 and as, per the example, the hammer handle is parallel to the ground, then gravity acts perpendicular to the hammer head mass, so, the linear acceleration of 10m/s^2 divided by the radius = 20 (radians/s^2)
So, remenber f=ma(linear) - Newton's second, well its the root of t=Ia(angular)
Torque is another name for rotational force or moment of force or moments.
Torque or Moment of force = moment of inertia or rotational inertial mass x angular or rotational acceleration
So, gravity accelerates 3kg hammer mass at 10m/s^2 which = 30N x lever arm 500mm = moment of 15Nm
The wrist and arm muscles must apply an equivalemt moment about the hammer handle but the force applied by the hand only has a small lever or moment arm. If the moment arm is say 50mm or 1/10th the moment arm available to the acceleration (force) of gravity acting on the mass of the hammerhead, then the force required is 10 times greater or 300N.
An example more relevant the podiatrist: when sprinting we want to move the legs as fast as possible, if the leg is extended then the inertial mass is further away from the hip pivot, say somewhere near the knee, and so the moment of inertia is high during swing phase compared to when the leg is fully flexed at the knee, ie bringing the foot toward the butt as the hip is flexed. In the latter case the leg is the same mass, obviously, but the moment of inertia is half way up the thigh and so much closer to the hip pivot, thus reducing the moment arm and so, for the same force applied by the muscles, the angular acceleration of the leg CoM will be proportionally greater.
If you think about it every movement involves levers and pendulums associated with Rotational motion of a mass. The leg about the hip, the lower leg about the knee the foot about the ankle, the hallux about the 1st MPJ, the whole body Centre of Mass travelling over the planted foot. While several bodies might have similar mass, the distribution of that mass has a great influence on the effort required by the muscles to overcome the inertia of the mass in rotation.
We don't need to use maths in our daily practice but I think it will help to see the maths to get a more intuitive picture that we can use daily to estimate or visualize what's happening to our patient.
So we've seen that a mass has inertia that needs to be overcome by a force to induce motion. In linear problems the inertia can be found by the equation of Newtons 2nd law f=ma - the inertia is proportionate to the mass x acceleration and in the opposite direction to the force applied.
In the case of a mass rotating about an axis though, the inertia or the resistance to rotational motion is expressed as moment of inertia (I) and I=mr^2 and mr^2 is expressed in kg.m^2.
The torque (t) due to the moment of inertia is a function of the acceleration of the mass and so we get the equation t = Ia (where 'a' is the angular acceleration in radians/second - not linear acceleration in meters per second) NB a radian is equivalent to the length of the radius of interest. Since the circumference of a circle is Pi (3.14) x diameter, there are therefore 6.28 (approx) radians in 360dgs of the circumference of a circle and one radian = 57.3dgs..
In the earlier example of the sledge hammer, most of the mass is distributed at one end of the long handle and it is possible to ignore the mass of the handle itself if we are not being too picky. So, in this case the equation I=mr^2 works directly but in the case of shapes where the mass is more evenly distributed then the change in moment of inertia is not so large when considering the torque at various axes relative to the CoM. If you go here you can see that the mr2 or mL^2 (where L = length instead of radius) of the various shapes and their rotational axes is multiplied by a fraction.
So if the hammer head has a mass of 3kg and the handle is 500mm and the acceleration is 20rad/s^2 then 3x(0.5*0.5)x20 = 15Nm.
If we assume for convenience gravity is 10m/s^2 and as, per the example, the hammer handle is parallel to the ground, then gravity acts perpendicular to the hammer head mass, so, the linear acceleration of 10m/s^2 divided by the radius = 20 (radians/s^2)
So, remenber f=ma(linear) - Newton's second, well its the root of t=Ia(angular)
Torque is another name for rotational force or moment of force or moments.
Torque or Moment of force = moment of inertia or rotational inertial mass x angular or rotational acceleration
So, gravity accelerates 3kg hammer mass at 10m/s^2 which = 30N x lever arm 500mm = moment of 15Nm
The wrist and arm muscles must apply an equivalemt moment about the hammer handle but the force applied by the hand only has a small lever or moment arm. If the moment arm is say 50mm or 1/10th the moment arm available to the acceleration (force) of gravity acting on the mass of the hammerhead, then the force required is 10 times greater or 300N.
An example more relevant the podiatrist: when sprinting we want to move the legs as fast as possible, if the leg is extended then the inertial mass is further away from the hip pivot, say somewhere near the knee, and so the moment of inertia is high during swing phase compared to when the leg is fully flexed at the knee, ie bringing the foot toward the butt as the hip is flexed. In the latter case the leg is the same mass, obviously, but the moment of inertia is half way up the thigh and so much closer to the hip pivot, thus reducing the moment arm and so, for the same force applied by the muscles, the angular acceleration of the leg CoM will be proportionally greater.
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