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Question

http://serc.carleton.edu/dmvideos/players/88129.html?hide_banner=true Watch the video to determine what is happening. Your task is to determine the coefficient of static friction between the man and the Graviton. Use the video experiment to determine the data for this result EXCEPT: There are 12 stationary red lines around the rotating platform. For the experiment, just when the man slides down the wall, assume that it took 12 frames of the video for one point on the rotating platform to go from one red line to the next. The frame rate of the camera is still 240 frames per second. What is the coefficient of static friction between the man and the wall?

Explanation / Answer

Hi,

In this case we have a body (the man) which is rotating in an platform. The forces over the man are the normal force, the friction force and the weight of the man.

In this problem we can use two axis, one radial (which can be called r) and the other vertical (which can be called as z).

r axis:   N = mar (there is no force that can counter the normal force)   

z axis: mg - f = maz (this represents the fall of the man)

As the movement has begun, the normal force and the friction force are related as follows:

f = uN ; where u is the friction coefficient (which is what we are looking for)

-The data in the video (at least what I could see) were:

The time to fall was from 480 frame to 530 frame which means that t was 0.208 s

The distance that the man falls is about 4.95 cm (which was calculated using the scale shown in the video)

The distance from the center of the platform to the man is about 28.5 cm (which again was calculated using the scale of the video).

With the previous data we can find the acceleration in the vertical axis:

z = (1/2)azt2     (as the man is initially at rest)

az = 2z/t2 = 2(4.95*10-2 m)/(0.208 s)2 = 2.29 m/s2

We could determine the radial acceleration (which is the centripetal one) using the video, but we will use the hint given:

If there are 12 stationary red lines, and it took 12 frames of the video for one point to go one red line to the next one, then the angular speed of the platform (an the man) is:

1 turn takes 12*12 frames, so the total time invested is 144/240 = 0.6 s (this is the period of the movement)

So, the angular speed is equal to: w = 2/T = 2/(0.6 s) = 10.47 rad/s

Therefore, the radial speed is equal to:

ar = Rw2 = (28.5*10-2 m)(10.47 rad/s)2 = 31.25 m/s2

in the end, we solve for the normal force, the friction force and finally the friction coefficient:

N = 31.25*m N

f = m(g - az) = 7.51*m N

u = f/N = 7.51/31.25 ::::::::::: u = 0.24

However, this is the coefficient of kinetic friction.

The coefficient of static friction is easier to calculate:

r axis:   N = mar (there is no force that can counter the normal force)   

z axis: mg = f (this represents the moment when the man is just about to fall)

The equation f = uN is valid again, but the value of u will be the coefficient of static friction.

f = 9.8*m N

N = 31.25*m N

u = 9.8/31.25 = 0.31

I hope it helps.

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