Objective:
The purpose of this experiment is to study the nature of Centripetal force by measuring it acting on a mass undergoing angular motion and centripetal acceleration. In this experiment we will evaluate relationship between Centripetal Force, mass, and velocity. Centripetal force is a force that acts on a body moving in a circular path which is directed toward the center around which the body is moving. The apparatus used for this experiment contains a vertical shaft fit into a bearing with a horizontal cross arm attached to the top of the vertical shaft (see Figure A). On one end of the horizontal cross arm there is a counterweight fixed in place and on the other end there is a mass (m), which is slung from a string and is attached
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to the vertical shaft by a spring. The shaft is then rotated; by applying torque with our hands. The faster we spin the shaft, the further away from the shaft the mass will move. By marking a particular distance that the mass is from the shaft, the number of revolutions per second can be observed and velocity calculated. The Centripetal force of the mass can be determined by attaching the mass it to a Newton scale and pulling it to the particular distance marked in the beginning. These calculations will be done numerous times to find relationship between Centripetal Force, mass, and velocity. Analyzing Centripetal force is important in objects traveling in a curved path. From atoms to planets and cars going around curved paths, centripetal force pulls the object to the centre. Knowing the concepts of how Centripetal force works can help us understand circumstances better. Theory: The experiment conducted is concerned with Centripetal force which is necessary to move around a curve. For this to occur, the object has to stay in motion around a circular path. We can recall Newton’s first law, stating that objects in motion want to stay in motion. In other words, there must be a force acting on an object to move it in a circular path so it does not move in a straight line. As well Newton’s second law, Force (F) is equal to mass (m) multiplied by the acceleration (a), is used in this experiment since there is an unbalanced force changing the direction of motion. When the mass is spinning around in a circle it is constantly changing direction undergoing acceleration. The force causing this change is the Centripetal force and it depends on the inertia and acceleration of the object. Therefore, we can use Newton’s second law to find the magnitude of the Centripetal force (Fc) which is equal to mass times the Centripetal acceleration (ac). Centripetal acceleration of an object moving in uniform circular motion is ac = v2/r. Knowing the magnitude of the centripetal force, mass and velocity (v) of an object circular orbit, the radius (r) can be isolated and the circumference around a circle can be found using 2πr. Therefore the equation for Centripetal force is Fc=(mv2)/r. Figure A1 Apparatus: Mass Pin Vernier Caliper Stop Watch Centripetal Force Apparatus Balance Ruler Procedure Measured the weight of the mass and record this value.
Used a Vernier Calipers and measured the diameter of the vertical shaft and record this value.
Placed the mass on the horizontal shaft. Moved the pin and mass 14 cm from the vertical shaft, and then fixed the pin in place below the mass using the setscrew. The distance was recorded.
Secured the mass to the spring the vertical shaft. Practiced spinning the shaft in order for the mass to travel in a circular path directly above the pin.
Once comfortable, timed how long it took for the mass to complete 25 full rotations for three separate trials. For each trial calculated the angular speed of the cross took the average of the three trials.
Removed the spring from the mass. Measure the centripetal force (Fc) on the mass by attaching it to a Newton’s scale and pulled the mass to the 14 cm mark above the pin. Read and recorded the value.
Repeated steps 2-5 above using the radii of 14.3 cm, 14.6 cm, 14.9 cm, and 15.2cm respectively.
Observations
Mass of Block: m=455.6g±.05g
Diameter of Shaft: d=1.13cm±.005cm
Trial r±.05 (m) R=r+½d (m) T25 (Time per 25 Cycles) ±.005 (s) T (s) w=2π/T (rad/s) ac
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(w2R) (m/s2) Fc ±.05N 1a .140 0.14565 25.61 1.024 6.134 5.479 2.5 1b .140 0.14565 25.64 1.026 6.126 5.467 2.5 1c .140 0.14565 25.53 1.021 6.153 5.514 2.5 2a .143 0.14865 25.33 1.013 6.201 5.717 3.0 2b .143 0.14865 25.29 1.012 6.211 5.735 3.0 2c .143 0.14865 25.35 1.014 6.196 5.708 3.0 3a .146 0.15165 25.10 1.004 6.258 5.939 3.4 3b .146 0.15165 25.02 1.001 6.278 5.977 3.4 3c .146 0.15165 25.07 1.003 6.266 5.954 3.4 4a .149 0.15465 24.94 0.998 6.298 6.135 3.7 4b .149 0.15465 24.88 0.995 6.313 6.164 3.7 4c .149 0.15465 24.79 0.992 6.336 6.209 3.7 5a .152 0.15765 24.53 0.981 6.404 6.465 4.2 5b .152 0.15765 24.44 0.978 6.427 6.512 4.2 5c .152 0.15765 23.81 0.952 6.597 6.861 4.2 Sample Calculations Time of One cycle, T, for trial 1a: T= T25/25=25.61/25=1.024s The angular speed, w, of the object which is calculated by the angle covered of 2π divided by the time taken, T, this is the sample calculations for trial 1a: w=2π/T = 2π/1.024=6.134rad/s The centripetal accelerations is calculated using the fact that ac=v2/R, where r is the radius of the path. Since v is the distance over time is can rewritten 2πr as the distance and T as the time, where 2π/T is w therefore the velocity can be written as v=w*r. Plugging that into the formula for centripetal acceleration you get ac=w2R. This is the sample calculation for trial 1a: ac=w2*r=(6.134)2*(.14565)=5.479 m/s2 Analysis The slope of the line is calculated by the change in y divided by the change in x: (w^2 r)/(w^2 rm)=1/m Therefore the slope of this line is supposed to represent 1/m where m is the mass of the block.
Conclusions
Percent error of the slope:
The slope of the line was supposed to be 1/m= 1/.4556=2.19 however the slope that was obtained was .8539.
%error= (theoretical value-measured value)/(accepted value)*100= (2.19-.8539)/2.19*100=61.0% error
The value that was obtained through the lab was much larger than the value of the slope that it was supposed to be.
There is a lot of human error that is not accounted for in this experiment as when the object is rotating the start time and end time must be manually started and stopped. Due to this the reaction time to start and stop may have been to slow or it may have been started or stopped too early. Furthermore for this experiment when the object is spinning the tangential velocity of the object may be changing because of the frictional force caused by the air and the friction of the shaft rotating on its axle. When the speed is reduced a little more force may be added causing overall variation on the velocity of the object thus affecting the time the object takes to do a revolution. Lastly when getting the object to begin the rotation, the alignment of the rotation of the object to the pin may have been
off. In order to improve the procedure of the lab you can increase the number of rotations that the object undergoes in order to reduce the error of the late or early timing. Another way to reduce error is to have a larger mass so the at the angular speed of the object will be less so that is easier to catch when the object passes the pin and whether the rotational path of the object was aligned with the pin. In the end though the value of the slope that was obtained was not exact to the accepted value, when taken into account that the in accuracies this lab demonstrated that measure of the centripetal force of the object and how that related to the angular speed of the object in motion by changing the radius of the motion.
The cup will stay on the plate throughout the entire rotation because it will be moving in a circular motion. We can see that moving in a circular motion will cause it to stay on the plate because of the equation v= ωr. This equation relates the angular velocity (ω) and the linear velocity (v). When the cup is placed at the very center of the plate the radius (r) will equal zero. When zero is put into the equation for r, the right side of the equation will equal zero, leaving us with the equation v=0. Because v is the linear velocity, we can see that the cup will not move in a straight line, rather a circular
[8.3] In what ways do you think your results would have been different if you had sampled at a different height on the rock?
One of the best methods for determining mass in chemistry is gravimetric analysis (Lab Handout). It is essentially using the the mass of the product to figure out the original mass that we are looking for. Thus the purpose of our experiment was to compare the final mass in our reaction to the initial mass and determine the change in mass.
UV-254 nm, 15 V, 60 Hz, 0.16 A). Masses were taken on a Mettler AE 100. Rotary
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When a mass is attached to the end of a spring the downward force the
from 10cm to 50cm to make it easier to see the difference in a graph.
I am going to begin by looking into going up in 0.1cm from 0cm being
This summer when you go to weigh that fat juicy watermelon, think about the mechanics of how the scale works. The basket is attached to a spring that stretches in response to the weight of the melon or other objects placed in it. The weight of the melon creates a downward force. This causes the spring to stretch and increase its upward force, which equalizes the difference between the two forces. As the spring is stretched, a dial calibrated to the spring registers a weight. When designing scales one needs to take into account that every spring has a different spring constant (k). Bloomfield (1997) defines k as “a measure of the spring’s stiffness. The larger the spring constant-that is, the stiffer the spring-the larger the restoring forces the spring exerts” (p. 82).
... spring, you are causing a twisting motion all the way down the coil. (Longhurst)
...e could add the mass piece without having them fall off. At the time of the experiment, this was not seen as a threat to our results.