Purpose: When objects move along circular paths a force is required to maintain this motion. This force is known as centripetal force and it produces an acceleration that we call centripetal motion to investigate the relationship between centripetal force and acceleration. We will also use our data to calculate the unknown mass of an object traveling in circular motion.

Equipment: Plastic tube,string,steel washers, rubber stopper,stopwatch,meter stick,triple beam balance,tape,paper clips, safety goggles.

Safety Precaution : be extremely careful when you swing the apparatus. Stay far away from classmates and object because this will ultimately result in someone getting hurt. Keep goggles on at all times to lessen the chance of eye injury.

Calculated Data:
Trial # Circumference (M) Period (s) Speed, v(m/s) Centripetal Acceleration (m/s²) Fc=Weight of Washer (N)
1 3.14 .88 3.6 25.92 .29
2 3.23 .65 4.5 39.32 .58
3 5.34 .74 7.2 61.33 .87
4 4.36 .60 7.3 76.68 .116
5 4.93 .56 8.8 98.65 14.30

Using the triple beam balance the mass of the rubber stopper was measured to be .0193 kg.

Measured Data:
Trial # # of washers Mass of washer (kg) Length of String
(m) Time for ten rev.
(sec)
1 5 .029 .5 8.79
2 10 .058 .515 6.46
3 15 .087 .85 7.14
4 20 .116 .695 5.96
5 25 .145 .785 5.58

Sample Calculations:
1. Circumference equals 2pr
We plug in the radius of the string, .5 and multiply
2p5=3.14 m

2. Period is the amount of time it takes to complete a cycle and since we measured the time it takes for ten we can divide that number by ten to get the period.
8.79 ¸ 10 = .87 second period

3. Speed equals 2pr¸T=2p.5¸.88=3.6 m/s

4. Centripetal acceleration is equal to v²¸r
3.6²m/.5=25.92 m/s²

5. Weight of washers is calculated by W=mg
W=(.029kg)(10 m/s²)=.29 N

6. Centripetal force=mv^2/r=(25.92)(.029)=.75 N

7.Slope: Dx/Dy=5.25

8. Actual rubber stopper weights .0193 , I measured 5.5 kg. Measured minus actual /actual times 100=percent error
Meaning I had a 500 percent error meaning I have done something very wrong.



Best fit line: y=73.9-33.9x
Slope=-33.9

GRAPH



Best fit line : y=27.21+5.25x
Slope: 5.25

Questions:
1.what is the direction of the stoppers velocity relative to its path? Explain
The direction of the stoppers is tangent to the circular path. This occurs due to the law of inertia in which states an object in motion tends to stay in motion. This is evident because if the string were to be cut the mass would fly tangent to the circular and go straight.

2.What directly supplies the centripetal force to the stopper ? In which direction is this force applied? Explain
The force is applied centripetically . This is supplied by the tension from the string attached to the mass .

3.Although the stopper is traveling with a constant speed it accelerates as it travels through a circular path. Explain why
Acceleration is a vector. It consists of a combination of direction and magnitude. The constant speed is the magnitude. The direction is changing constantly because it is in a circular path. Therefore, it is accelerating.

4.How does the centripetal acceleration vary from trial to trial? Explain. What is the direction centripetal relative to the stopper’s path?
The centripetal acceleration slowly increases . This occurs because the velocity increases as more weights are added . Centripetal is towards the stopper because it is near the center of the apparatus and centripetal mean center seeking .

Conclusion :
According to Newton’s first law, an object in motion continues to move with a constant velocity in the same direction unless acted upon by an external force.