By Erin Krysinski
My mother recently underwent surgery to correct a herniated disc that had ruptured in her neck. Though she has had neck problems for years, my father, as per usual, blamed the rupture on a particular incident that occurred while my mother was riding our trusty thoroughbred. My father has taken a liking over the years to blaming most things on the horse, money, injuries, etc, probably because horseback riding is a rather expensive and dangerous hobby. The event occurred as follows: My mother was cantering into a fence, and misjudged the distance. While she continued forward, the horse decided to add in another stride for safety reasons, and in doing so, significantly reduced his speed within a very short amount of time (deceleration). My mother as a result, felt a sort of whip lash as she quickly was forced to decelerate with the horse beneath her (see these other examples: http://www.youtube.com/watch?v=DvJVTHk2m74 ). Of course, various other incidents occurred that same week, including lifting heavy dogs (my mother is a veterinarian), and moving wheelbarrows at the farm, so it was difficult to pinpoint which incident was the herniation perpetrator. My mother and even more so, my father, seem convinced that that horse jumping incident was the cause. I on the other hand, would like to prove otherwise. I hypothesize that a single incident of whiplash in this case, is not capable of rupturing an intervertebral disc.
In order to determine whether the horseback event was capable of causing the rupture, I first looked at the force on my mother when the horse quickly decelerated. To find this force, I used the impulse equation, ∑Fx=∆p/∆tà (m(v2-v1))/∆t=∑Fx , to find the sum of the forces on my mother in the x direction. I estimated that as the horse added an extra stride, and thus almost stopped, decelerated from v1= 6.94 m/s (an average canter speed) to v2=1.39 m/s (an average walk speed) in t=0.25 sec. The mass of my mother + the horse= 1115 lbs, or 505 kg. If we plug this into the equation, (505kg(6.94m/s-1.39m/s)/.25sec =∑𝐹. In this case, ∑F=11,211 N in the x direction.
To find whether this force is enough to herniate her disc, we need to first find the ultimate shear strength of a intervertebral disc, and then find the minimum force that would cause a disc to rupture. From “Spine Biomechanics” by Rapoff, we estimated that the ultimate shear strength of an intervertebral disc is about 17.6 N/1mm or 1.76x107N/m2 . To find the minimum force that would break a disc, we can use the equation: ultimate shear strength=F/A. To find the area of the disc, we use the average disc radius (.02m, estimated from http://arthritis-research.com/content/5/3/120) and then use the circle area equation (A=2πr2) and then get that A=0.00126 m2. From here, we can plug in our ultimate shear strength and our area to the above equation to find the minimum force applied for rupture. Therefore, 1.76x107N/m2=F/0.00126. F=22,116.81 N. So therefore, the shear force would have to be greater than 22,116.81 N to herniate the disc. Because our F is only 11,211N, it’s not likely that the horse event alone ruptured the disc, rather a series of events over many years broke down it’s shear strength, until a smaller event could
In reality things are a bit more complicated than this. We would need to break the force on the rider down further and find what force is actually applied to the disc itself. We can do this by looking at the neck as a rotating rod with a sphere on top, and can view the deceleration as a torque applied to the seat of the rider, since that’s where the deceleration occurred. Because we know the deceleration of the horse, we could then use atan=rxalpha to find radial acceleration, and then use kinematics to find radial velocity as well as theta. We could also find omega via w=v2/r and then use kinematics to find theta. Regardless we can then use the conservation of energy equation, Wnc=∆KE+∆PE, or in this case, ∆1/2 I w 2 +∆1/2mv 2 + ∆mgh= (torque)(theta) to find the torque applied. With torque, we can then use Torque=rFsin(theta) to find the force applied. Another way to find the force applied to the disc, would again be the conservation of momentum theorem, this time rotational. If we calculated I for the person on the horse as a rod with a sphere on the top and consider the axis of rotation their seat, after finding w1 and 2 as illustrated above, we can then use Iw1=Iw2 to find the change in rotational momentum, ∆L. Because ∑torque=∆L/∆t, we can then find the sum of the torques, which in this case is only one at the seat, and then as illustrated above, find the force applied via torque=rFsin(theta). We’ll save these calculations for another time, but since the total force of the system wasn’t enough to rupture the disc, we’ll also assume that the torque on the body alone won’t be enough either.