Person Sliding on IceBy Matt Vonk

1. Carefully watch the Person Sliding on Ice video. What are a few things that you notice?

There’s a shovel in the snow. I wonder if they shoveled off the ice first.

There’s the car from the other video parked on the side.

His son is making the “he’s crazy” hand gesture

Someone got hot and left their jacket on the goal

2. How fast is the person going when he starts to slide?

The middle of his body seems to cross -4 m at frame -52. The middle of his body seems to cross 0 m at frame 0.

v = displacement / time = 4 m / 52 frames x (60 frames / 1 s) = 4.62 m/s = 10.3 mph

3. How far does he travel sliding?

6.85 m

4. How much time does it take the person to stop, once he starts to slide?

His shoes seem to stop sliding at frame 139.

5. What is the average acceleration while the person is sliding?

Acceleration = (vf – vi)/time = (0 – 4.62) / (139 – 0) x (60 frames / 1 second) = 1.99 m/s2

6. Average velocity can be found two ways. Assuming that the acceleration is constant, solve for the average velocity (for the time that the person is sliding) using both methods.

a. vavg = displacement/time

6.85 m / 139 frames x (60 frames / 1 second) = 2.96 m/s

b. vavg = (final velocity +initial velocity)/2 NOTE: This equation is valid when the acceleration is constant.

(0 + 4.62 m/s) / 2 = 2.31 m/s

UncertaintyThe two numbers that I got above when I calculated the average velocity are not exactly the same. Yet, we can still ask if the two values agree, that is, do the range of values for the two numbers overlap when we consider the uncertainty of each number. All measurements have some uncertainty. In this case there is some uncertainty in the measurement of distance and time.

7. Estimate the uncertainty in your displacement measurement. Express your answer in meters.

0.05 m

8. Estimate the uncertainty in your time measurement. Express your answer in frames and in seconds.

1 frame or 0.017 seconds

9. Recalculate vavg above using both methods, but this time add or subtract your uncertainty to each factor (displacement and time) to find the largest spread possible. Using equation 1 above, vavg will be the highest when the displacement is large and the time is small.

For example if previously you had found that the person travelled 7 meters in 52 frames your value for vavg would have been:

(7 m / 52 frames) x (60 frames / 1 second) = 8.08 m/s

If you think your displacement uncertainty is 0.1 m and your time uncertainty is 2 frames then you would re-solve for vavg in a way that produce the most spread, i.e.,

(7 + 0.1 frames ) / (52 – 2 frames) x (60 frames / 1 second) = 8.52 m/s this is your HIGH value

(7 – 0.1 frames) / (52 + 2 frames) x (60 frames / 1 second) = 7.67 m/s. this is your LOW value vavg high value using method 1 (6.85+.05)/(139 – 1 frame) x (60 frames / 1 s) = 3.00 m/svavg low value using method 1 (6.85 -.05)/(139 + 1 frame) x (60 frames / 1 s) = 2.91 m/s

vavg high value using method 2 (0 + 4 + 0.05m) / (52 – 1 frame) / 2 x (60 frames/1s) = 2.38 m/svavg low value using method 2 (0 + 4 - 0.05m) / (52 + 1 frame) / 2 x (60 frames/1s) = 2.24 m/s

10. When you compare the ranges of vavg using equation 1 and equation 2 above, do the ranges overlap?

In this case the ranges don’t overlap. That probably means that there is something else going on than just measurement uncertainty. To figure out what was going on I made a graph of his position over time.

Looking at the graphs I notice that the person didn’t slow down very much for the first ¾ of a second (the first 3 meters that he slid). After that the person decelerated at a fairly constant rate (See graph of “Last 9 points…” above).

When you calculate average velocity using the first equation, (vavg = displacement/time) it doesn’t depend on whether or not the acceleration was constant or not. But the second equation (vavg = (final velocity +initial velocity)/2) only works if the acceleration is constant, i.e., if the velocity vs. time graph is a straight line. Looking at the upper graph above, it seems clear to me that the points do not fall on a line and that the acceleration was not constant. Rather, it looks like there was very little acceleration for the first 3 meters (~0.75 s), and a greater acceleration at the end. This suggest that there was actually less friction on the slider when he was moving quickly than when he was moving slowly. It seems like the ice has a velocity dependent coefficient of friction, and that it’s actually more slippery the faster you go. It would be fun to pull a weigh over ice with a force probe and actually test that.

It’s interesting to look back at the results of both methods. The first method resulted in a larger average velocity value than the second method. That’s because the slider went farther (greater displacement) than he would have if the acceleration had been constant and he had started slowing down immediately.

It looks like the person doesn’t slow down much in the first ¾ of a second.

Forces & Friction

11. Why is the person slowing down?

The ice is exerting a force on him in a direction that is opposite his velocity.

My guess is that some of you said that the person is slowing down because they stopped running. That’s a good step, but in physics class we’ll think about it in a little different way. Newton’s first law says that nothing can change its velocity unless there is a net force on it. Note: The net force on an object is just the sum of the forces on an object. If you add up all the forces on an object and the sum is zero then the object will maintain its velocity (both speed and direction).

12. What object outside of the person is forcing him to stop? What is the direction of that force?

The ice exerts a force on him toward the right.

13. What is the coefficient of friction between the person’s boots and the ice?

fx = Ffriction = n = (mg) = max → m=ax/g = 1.99 / 9.8 = 0.20 (dimensionless)

14. What is the magnitude of the frictional force if the person has a mass of 90-kg?

Ffriction = mg = (90 kg)* (9.8 m/s2) * 0.20 = 176 Newtons! Or about 40 pounds.

15. What could you change about the situation to increase the frictional force on the person?

Increase the normal force by having the person carry additional mass.

Increase g by moving to a heavier planet or performing the experiment in an underground mine (equipped with an ice rink)

Increase , by roughing up the ice or by using boots with more traction.

It’s interesting to note that having the person carry more mass would increase the frictional force on him, but it wouldn’t decrease the stopping distance or time. That’s because the added mass would also have inertia.