The physics of circular motion sets an upper limit to the speed of human walking. (If you need to go faster, your gait changes from a walk to a run.) If you take a few steps and watch what's happening, you'll see that your body pivots in circular motion over your forward foot as you bring your rear foot forward for the next step. As you do so, the normal force of the ground on your foot decreases and your body tries to 'lift off' from the ground. A person's center of mass is very near the hips, at the top of the legs. Model a person as a particle of mass m at the top of a leg of length L. Find an expression for the person's maximum walking speed v_max.

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Hi, everyone in this practice problem, we are being asked to determine the minimum speed at which a pedestrian will start to run. We will have a motion capture system employed to track a motion of an adult pedestrian walking over a rigid level surface and the position of the center of gravity of the pedestrian is going to be investigated. According to the data analysis, the pedestrian center of gravity follows a vertically curved or circular path when walking and the center of gravity is defined by an arc with a radius of one m. We're being asked to determine the minimum speed at which a pedestrian will start to run. And the options given are a 2.5 m per second. B 3.1 m per second, C 6.2 m per second and D 9. m per second. So we'll model the pedestrian as a point like particle with its mass concentrated at the center of gravity. The center of gravity of the pedestrian is going to be described by a circular vertical motion of radius one m, which is given in the problem statement. So the center of gravity is then going to be subjected to two different forces. First is going to obviously be the weight which is going to be pointing vertically downward. And the second is going to be the normal force which is going to be pointing vertically upward. So the first one is the weight which is going to be indicated by this M multiplied by G. And the second one is going to be the normal force pointing upwards. The weight is pointing downwards which is just going to be end. OK. So we wanna apply Newton's Second Law at the highest along the radial direction, which will actually give us Sigma F R equals M multiplied by A R. And the only two forces are M G and N and I'm going to um take the convention where M G is going to be positive. So this Sigma F R is then going to be M G minus N equals to M multiplied by A R. You want to recall that the radial acceleration A R is going to equals to or can be also fine by V squared divided by R or the radius of the four of the motion. And in this case, you want to substitute the V squared divided by R into our Newton Second Law equation so that we get M G minus N equals to M multiplied by V squared divided by R. So the minute of the normal force may actually vary from zero to a maximum value of M G. And when the normal force is zero, we will then obtain the transition speed or V limit from walking to running. So the normal force here can vary from zero to M G. And we limit will happen when it or the pedestrian is transitioning from walk to run. And that means our normal force will have to be zero. So then our equation employing this condition will then equals to M G equals to M multiplied by V limit squared divided by R just like. So, so if we want to find this transition speed, we do have to replace our normal force with zero Newton. And in this case, we found our formula right here which we can then simplify because we can cross out the M so that we get an equation for V limit equals to the square root of G multiplied by R. So the R is given in the practice or in the problem statement which is the radius of one m. So then we can actually solve for V limit, which will then be equals to the square root of G which is 9.81 m per second squared as usual multiplied by R which is going to be one m. And that will give us the fe limit value of 3.1 m per second. So the minimum speed at which the pedestrian transitions from walking to running or V limit is actually going to equals to 3.1 m per second, which will correspond to option B in our answer choices. So option B is going to be the answer to this particular practice problem and that'll be it for this video. If you guys still have any sort of confusion, please make sure to check out our adolescent videos on similar topics and that'll be it for this one. Thank you.

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Key Concepts

Circular Motion

Center of Mass

Normal Force