Uniform Circular Motion Physics 201L lab report PDF

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Uniform Circular Motion PHYS 201L-014 10/25/2020 Introduction The objective of this experiment was to use uniform circular motion which, occurs when an object is moving in a circle where the radius and the speed is constant. Although radius and speed are constant, there is a force still acting on the object called centripetal force which pushes against what the object would normally do since, it would either sit still or travel in a straight line. From there, you can use the centripetal force equation to find the mass of an unknown object. Procedure 1. We started without weighing the bob on the scale, we placed a piece of masking tape on the radial measurement that we wanted to calculate. 2. To start, you must increase the speed until the bob is hitting the tape, meaning its on the correct radius. 3. You start timing the speed of the bob as soon as it reaches the masking tape and stop the timer once you hit five complete rotations. 4. Record the time that it took down on your table and repeat for four more trials 5. Find hanging mass by adding weight until the string is stretched out to the radius that you previously measured and record the mass Repeat these steps for the other four radii that you decided to work with and record the time for each trial and the hanging mass for the four radii. Diagrams

Radii (m)

time (s)

t2

t3

t4

t5

Mass (kg)

T

T2

ω

a (m/s2)

Weight (N)

0.14 0.16 0.18 0.20

11.01 9.62 8.62 7.96

11.31 10.11 8.70 8.12

12.11 9.87 9.18 7.82

11.55 9.44 7.49 8.36

11.10 9.48 8.09 7.62

0.316 0.449 0.683 0.989

11.416 9.704 8.416 7.976

2.2832 1.9408 1.6832 1.5952

5.213 3.7667 2.83316 2.54446

2.75192 3.23742 3.73288 3.93881

1.06023 1.67694 2.50819 3.10284

3.09996 4.40469 6.70023 9.70209

0.22

7.00

7.08

6.87

7.77

7.86

1.131

7.316

1.4632

2.14095

4.29414

4.05672

11.0951

Data tables Analysis The mass for each trial was calculated to equal 2.92386, 2.62662, 2.67134, 3.12684 and 2.735. Since we did not weigh the bob in person, the expected value is the average of the individual masses which is equal to 2.81673 kg. The percent uncertainty is equal to 1.458% which means that the experiment was relatively successful since the two masses found do not differ by much, given the fact that the bob was not weighed in person and user error during data collection can affect the data set and final calculations. The data shows that weight and acceleration are directly proportional since, when the weight goes up, the acceleration goes up as well. T ¿

( 11.416 ) =2.2832 ( 5)

ω =

2π 2π = =2.75192 T 2.283

a=rω 2= (0.14 )(2.75192)2=1.06023 m/s2 µg ( 0.316 kg )( 9.81 ) m v2 =m= 2 = =mass=2.92386 kg r rω 1.06023 m / s2

F=

2 Weight ( N ) =( 0.316 kg ) ( 9.81m/s )=3.09996 N

Uncertainties T= ω= a=

2.2832+1.9408 … =1.79312= ( 2.2832 −1.79312 )+ (1.9408 −1.79312 ) +… =± 0.255104 5 5

0.839114 + 0.353614 +… 4.29414+ 3.93881 + … =±0.477091 =3.59103= 5 5

1.42075 + 0.804044 +… 4.05672+ 1.06023 … .. =± 0.88992 =2.48098= 5 5

Graphs Weight versus acceleration

Weight (N)

rω2=a (m/s2) y=mx +b — F=( slope ) a+( y intercept=0 )

slope =mass =2.84589 kg

percentuncertainty=

2.84589−2.81673 =0.01458=1.458 % 2

Questions 1.) When accelerating a car traveling in a straight line, you will initially feel an increase in friction from your tires against the road and the increase in wind resistance against the car but, this decreases as you continue accelerating and the force only increases if you start slowing down. When you are turning sharply in a car, the change in direction makes it feel as if you are going faster when you did not accelerate at all. 2.) The magnitude of the force is F=mac with a being centripetal acceleration since, it is moving in uniform circular motion and the force is going to be pointed inward towards the center of the apparatus....