O n l i n e E xp e r i m e n t - C e n t r i p e t a l F o ...

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PHYS 191 - Physics I Labs

Online Experiment - Centripetal Force Line 1: Name of the experiment? (Centripetal Force)

Line 2: Your full name: (last, first)

Line 3: Lab course? (Phys 191)

Line 4: Section? (two digits code)

Line 5: Purpose of the experiment? (find the purpose after reading the theory and procedures)

Theory The circular motion is an important case of 2D motion since many physical systems exhibit rotational motion. In this writeup we will first investigate the kinematics of the uniform circular motion and then we will formulate the mechanics of it later.

As shown in the figure >>> imagine an object with mass that is moving on a M circular path with radius such that the R angle swept by the position vector with r→ respect to the +x axis (i.e. the trigonometric angle) is a linear function of time. Hence we can write that where (t) tθ = θ = θ0 + ω

is a constant and is the initial angle ω θ0 (i.e. the angle at time t=0). The derivative of the angle function (the angular position) will give us the angular speed: . θ dt d / = ω

Therefore is the constant angular speed in ω this formulation. The object swipes the same angle for all the same time intervals. This is why this motion is called “uniform” circular motion.

Now let’s express the position vector in terms of by using the unit basis vectors and : θ i j Eq1: (θ) i (θ) jr→ = R cos ˆ + R sin ˆ

If we take the derivative of the position vector we can obtain the velocity vector:

Eq2: R (θ) i R (θ) jv→ =− ω sin ˆ + ω cos ˆ

Prepared by the lab supervisor, Dr. A�lgan (ea�lgan01@manha�an.edu) Online 191-Centripetal Force PAGE 1 / 8

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This vector is tangent to the circle and oriented in the direction of rotation, as we intuitively expected. The magnitude of the velocity (i.e. the speed of the object ) is also a constant: v

R R v = √[ R (θ)] ωR (θ)]− ω sin 2 + [ cos 2 = ω √sin (θ) (θ)2 + cos2 = ω so

. R v = ω

This is an important relation in rotational dynamics that connects the linear speed to the rotational speed . If we take the derivative of v ω the velocity vector (it is in Eq2), we can obtain the acceleration vector:

Eq3: R (θ) i R (θ) ja→ =− ω2 cos ˆ − ω2 sin ˆ

Note that , so the [R (θ) i (θ) j] ra→ =− ω2 cos ˆ + R sin ˆ =− ω2→ acceleration vector is in the opposite direction to the position vector and since the position vector is always radially outward, the acceleration is always towards the center of the circle (if it is placed on M, of course), i.e. it is centripetal. The magnitude of the acceleration is also uniform and given by

. a∣ r∣ ∣r∣ R a = ∣→ = ∣ − ω2→ = ω2 → = ω2

Since , and inserting this last expression into the Rv = ω R ω = v/ above equation gives us

. R a = v2/ In short, in uniform circular motion the acceleration is centripetal and its magnitude is equal to . So far we have not used any R v2/ physical law. These results are the consequence of simple algebra.

Now let’s apply Newton’s second law of motion onto the uniform circular motion. Since , the net force is always in the same aF

→net = m→

direction with the acceleration which is centripetal in this case. For this reason, in physics, the net force in uniform circular motion is called “centripetal force”. The magnitude of the centripetal force, let’s denote by , therefore is equal to F c

. v R F c = M 2/ In this lab we will justify this result on a rotating bob system. The linear speed in uniform circular motion is where is πR T v = 2 / πR2 the circumference of the circle, i.e. the distance taken in one cycle and is the period of the motion, i.e. the time that one cycle takes. T By inserting this expression into above equation we can also write

. π MR T F c = 4 2 / 2

In this last expression , , and are measurable quantities in M R T our experimental setup as you will do in section Procedure. The newtonian dynamics predict that the centripetal force ought to be equal to . Hence we can calculate it by measuring , π MR T 4 2 / 2 M R and . The setup also allows us to measure the net force acting on T

“directly” by using the stretch of a spring so that we can compare M those two quantities to justify the validity of our prediction.

NOTE: For general cases, (i.e. if the motion is a non-uniform circular motion), the centripetal force is defined as the net force in the radial direction only, but has the same properties with those of the uniform circular motion (i.e. cenrtipetal and equals to ). R Mv2/


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Procedure Step 1. The experimental setup. The experimental apparatus is a very simple table

top “crane” where a bob with mass is hanging from the crane’s arm rod with M a string and also attached to the axle with a horizontal spring on one side. On the other side, a hanging mass is temporarily connected to with a string m M as shown in the figure >>>. In this configuration the spring is stretched , and

is aligned with the indicator. The picture of the apparatus is under the M Downloads tab at https://mc-phys-labs.weebly.com/ and named as “centripetal-force.jpg”. Click and see the picture (you do not need to download it). By using the information given in the picture answer these following questions:

Line 6: M = ________ kg.

Line 7: m=________ kg.

Line 8: What is the magnitude of the force applied by the hanging mass onto M through the tension of the string? ________ N. m

Line 9: What is the magnitude of the force applied by the spring onto M in this stationary configuration? _________ N.

Line 10: What is the direction of the force applied by the spring onto M in this configuration? _______ [ left | right | up | down ].

The same crane can be seen rotating in movie centripetal-force-25fps.gif ( under the Downloads tab). Click onto the movie link ( it says “Download File” ) and watch it in your web browser. Note that the hanging mass is detached m


https://mc-phys-labs.weebly.com/

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from and the system is rotating at such a speed that the bob is again aligned with the indicator when passing over it. The alignment is M not perfect but you should assume that the bob is aligned perfectly to answer the rest of the questions correctly.

Line 11: The spring in the rotating system is stretched by the same amount as in the stationary case. _______ [ true | false ].

Line 12: What is the magnitude of the force applied by the spring onto M while the system is rotating? _________ N.

Line 13: What is the direction of the force applied by the spring onto M while the system is rotating? ________ (give your answer in line with the terminology that is used in the theory section)

Line 14: What is the magnitude of the net force acting on while the system is rotating? ________ N M

Line 15: What is the magnitude of the centripetal force acting on then? ________ N. (Remember, as we stated in the theory M section, the centripetal force is the net force in uniform circular motion)

Step 2. How the movie is created. The rotation of the crane is recorded on a smartphone camera. By using an image processing program the

size of the movie is reduced and uploaded onto our lab website with name centripetal-force-25fps.gif . The frame rate of the movie is 25 fps. Go to our lab website https://mc-phys-labs.weebly.com/ and click the Downloads tab and download the movie. You need to right click on the image while it is playing and then select “save image as” to download.

Line 16: What is the time that elapses between two consecutive frames? _______ s

Step 3. ImageJ. You will find the period ( ) and the radius of the rotation ( ) by analysing the movie by using a free image processing T R program from NIH named “ImageJ”. If you do not have ImageJ, Go to https://imagej.nih.gov/ij/download.html , download, install and launch it. This control panel should show up (be patient, might take more than 5 seconds to appear):


https://mc-phys-labs.weebly.com/

https://imagej.nih.gov/ij/download.html

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Step 4. Set the play speed to the real time speed. Open the movie (File>Open> select the movie file) and set the animation speed to 25 fps:

Image>Stacks>Animation>Animation Options and enter 25 for the “speed” in the window. If the speed is already 25 just hit OK. At these settings the movie will play at real time speed. This means 25 frames will pass in one second.

Step 5. Set the distance scale. The length of the ruler that you see in the movie is actually 50 cm. Use + and - keys (or magnifying glass tool as shown in the figure >>>) to maximize/minimize the image. You can zoom in/out where your pointer is. Take advantage of this feature often in order to align the measurement tools of the program with better accuracy.

Now select the line segment tool as shown (the button that is clicked). >>>

Align it along the length of the ruler as shown. By zooming in and out (+ and - keys) you can fine tune your alignment.


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Go to Analyze > Set Scale. A window will open. Enter 50 into the “Known distance” box, and enter cm into the “Unit of length” box and click OK. This will set the scale for the positions and distances measured. All distance and position measurements will be the actual, real life values in units of cm! >>>

<<< Now measure the length of the crane’s base as shown.

You can read the length in the “communication bar” at the bottom of the control panel >>>. The length info appears only if you hold one of the line segment points with your mouse and move it . The base length is around 42.5 cm. Your measurement should be close to 42.5 cm within an error of 1 cm. if not, you could not set the scale correctly! Repeat this step again.


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Step 6. Measure the radius . Measure the radius of the rotation at frame 5 by R

spanning the line segment tool from the center of the axle to the center of the bob as shown >>>

<<< The index of a frame is always shown in the movie window, at the top left corner of the image. In this example, 5/39 means there are 39

frames in the movie and the 5’th frame is currently on display.

Line 17: What is the radius? = ________ m (read it on the communication bar, do not forget to convert cm to m) R

Step 7. Measure the period . If you play the movie (with no cyclic repetitions) you will notice that the bob passes over the indicator only T twice. If you need it (in fact, you will do :-P), you can pause the movie at any time and can go back- or for-ward, one frame at a time, by using the arrow head buttons at the bottom of the movie window at the edges of the sliding bar. If you are using a mouse then the wheel of the mouse will do the same job for you if you roll it up or down. MAC users may not see the back- or for-ward buttons, in this case shift-left or -right on the keyboard does the same. Find at what frames the bob aligns with the indicator.

Line 18: Frame index of the first alignment is = ________ . F 1

Line 19: Frame index of the second alignment is =________ . F 2

If you noticed, from the first alignment to the second one the bob rotates only once! Therefore by using the number of frames passing between the two, we can calculate the period.

Line 20: =________ . F 2 − F 1


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Line 21: =________ s. (you need to use line 20 and line 16 to do the calculation) T

Line 22: Calculate the centripetal force =_______ N . (See the theory section for the formula. You will need to use the values F c in line 6, 17 and 21)

Line 23: Compare the value you found in line 22 to the one found in line 15. Give your answer as the percent discrepancy between the two by taking the line 15 value as your reference: _____ %

Line 24: State your conclusion.


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