Objective
To measure the centripetal force for whirling a mass round a horizontal circle and compare the result with the theoretical value given by F = m(2r .
Apparatus
12 slotted weights with hanger (0.02kg each)
1 rubber bung with nylon string about 1.5m
1 glass tube about 20cm long
1 triple beam balance
1 meter rule
1 stop watch
Several small paper markers
Theory
When a mass m attached to a string is whirled round a horizontal circle of radius r, the centripetal force for maintaining the circular motion is given by F = m(2r , where ω is the angular velocity of the circular motion. The centripetal force is provided by the tension of the string.
The string will always make an angle( to the horizontal instead of lying on the plane of the horizontal circle described by the mass m. Thus, the centripetal force is given by the horizontal component of the tension. It is shown that the tension T = m(2L regardless of the angle(.
Procedure
1. Measure the mass m of the rubber bung by the triple beam balance.
2. Attach the rubber bung with nylon string thread through the glass tube and a number of weights. First start with M = 0.12kg.
3. Measure the length of L from the rubber bung to the glass tube (i.e. 0.8m). Adjust the position of a small paper marker 1cm below the glass tube. The set up is shown below.
4. Hold the glass tube vertically and whirl the rubber bung around above your head in a horizontal circle. Increase the speed of the rubber bung gradually and allow it to move out (i.e. let L increases) until the marker is about 1 cm below the lower end of the glass tube.
5. Keep the angular speed constant so that the marker is always about 1 cm below the glass tube throughout. Ask your partner to time 20 revolutions of the bung using a stop watch. Start the stop watch at 0 and stop it at 20. Take one more confirmatory reading and obtain the mean time for 20 revolutions.
6. Calculate the angular velocity ( and hence find m(2L.
7. Repeat the experiment using different masses M (i.e. 0.16kg, 0.20kg, 0.24kg).
Precaution
1. Make sure that the surrounding is cleared before the experiment.
2. You should wear safety goggles.
3. The glass tube should be held above your head.
4. The angular speed of the motion should be kept constant so that the marker was always about 1 cm below the glass tube throughout.
Result Mass of the rubber bung m = 0.0543 kg
|M / kg |Mg / N |Time for revolutions 20t / s |Angular speed |m(2L / N |
| | | |[pic][pic] / rad s-1 | |
| | |1 st trial |2 nd trial |Mean | | |
|0.12 |1.2 |22.3 |22.1 |22.2 |5.66 |1.39 |
|0.16 |1.6 |21.8 |21.7 |21.8 |5.78 |1.45 |
|0.20 |2.0 |19.0 |19.1 |19.1 |6.60 |1.89 |
|0.24 |2.4 |17.4 |17.1 |17.3 |7.28 |2.31 |
Discussion
1. The string is not exactly horizontal when the rubber bung is whirling, but the angle( made between the string and the horizontal is very small, so the tension T is calculated as m(2L regardless of the angle(, the values of m(2L are similar to the values of m(2r .
The centripetal force is given by the tension in thread, because T = Mg, so the centripetal force is also equal to the weight of the slotted weights (i.e. F = Mg). From the result above, the values of the centripetal force are similar to the values given by F = m(2L (i.e. Mg= m(2L).
2. The sources of errors in the experiment There are two error in this experiment, which are systematic error and random error. The systematic error is the reaction time of the person who starts and stops the stop watch, this will affect the time needed. The random error is the air resistance during the experiment, this will increase the time needed for the revolutions.
3. Improvement The experiment should be done by the same person. Also, the number of revolution should be more so that the systematic error caused by the reaction time will decrease.
Conclusion
From the result, the centripetal force increases when the weight of the slotted weights increases. Because of the problem and errors stated above, the values of the centripetal force is similar but not exactly the same with the theoretical values given by F = m(2r .
-----------------------
mg
L
T
(
r
mass m
r
F
Rubber bung
Slotted weights with hanger
M
Small paper marker
Glass tube
Nylon string
L
m
Glass tube
Small paper marker
M
M
Small paper marker
Glass tube
m
L
(
Mg
T
T
You May Also Find These Documents Helpful
-
1 Draw a cross on a piece of paper and put it underneath a 100 cm3 measuring cylinder, or use the cross from part 2, so it can be seen when looking down the cylinder.…
- 735 Words
- 4 Pages
Satisfactory Essays -
8. Holding your finger over the hole in the rubber stopper, invert the tube into the 400-mL beaker of water. Clamp the gas collection tube in place on the ring stand so that the stoppered end of the tube is under the water but not touching the bottom of the beaker.…
- 910 Words
- 4 Pages
Better Essays -
4. The net force acting on a body in uniform circular motion is constant in…
- 514 Words
- 3 Pages
Satisfactory Essays -
1. Fill a shallow pan 3/4th full of water and place the glass tube into the water and make sure that the tube is covered with water and has no air bubbles.…
- 1242 Words
- 6 Pages
Good Essays -
3. Place the lass on the table at the starting point. Hold the force meter parallel to the tabletop. Pull the mass at a constant speed along the surface of the table to the 0.50 m mark. Observe the force measured on the force meter…
- 330 Words
- 2 Pages
Satisfactory Essays -
3. After setting up the equipment place a ring of artery from a hook on a clamp stand.…
- 789 Words
- 4 Pages
Satisfactory Essays -
multiply this reading by 0.01. If the head is at or immediately to the right of the half mm graduation, add 0.50 mm to the reading on the rotating scale.…
- 166 Words
- 1 Page
Satisfactory Essays -
Newton’s laws of motion govern the motion of an object in an inertial reference frame which is one which is not accelerating. As a rotating reference frame is constantly accelerating, a centrifugal force must be introduced in order for Newton’s Laws of Motion to apply. Centrifugal force is a useful concept in analysing behaviour in rotating systems and is the apparent force acting outwards to counteract the centripetal force. This apparent force results from an objects tendency to continue in a straight path due to its inertia.…
- 3393 Words
- 14 Pages
Best Essays -
Step 3: Calculated the length of the spring without any external force added to it.…
- 330 Words
- 2 Pages
Satisfactory Essays -
Centripetal force is the centrally directed net force acting on the body. It provides the centripetal acceleration which keeps the body moving in a circle according to Newton’s Second law (Fnet=ma). In this experiment, the tension is provided by the weight of screw nuts and the centripetal force is supplied by the tension. The tension in the string is resolved into two components(Tsinθ and Tcosθ).As there is no vertical motion, the vertical component of tension is balanced by the weight of the rubber bung (i.e. Tcosθ=mg).The horizontal component of the tension provides the net centripetal force (i.e. T=mLω2). By comparing the two values of tension, the expression of centripetal force (Fc=mrω2) can be verified…
- 1298 Words
- 6 Pages
Powerful Essays -
Since the hanger mass is 50g, it is to be added to the mass of the weight.…
- 546 Words
- 3 Pages
Good Essays -
d. to conduct and experiment to determine the effects of the two factors on the rate of…
- 1264 Words
- 6 Pages
Good Essays -
Edexcel AS/A level Mathematics Formulae List: C1 – C4, FP1 – FP3 – Contents Page – Issue 1 – September 2009…
- 3379 Words
- 14 Pages
Good Essays -
For that experiment, we use the conservation of energy equation to find the moment of inertia of the cross arm, and then use the definition equation of the moment of inertia to get the exact magnitude for the cross arm, the disk and the ring.…
- 715 Words
- 4 Pages
Satisfactory Essays -
Procedure First we measured the weights of two masses and wingnuts that secure them. Then we placed one of the masses on the very end of a horizontal rod on the centripetal force apparatus, 0.162 m away from the centre of the rod, and the other mass 0.115 m away from the centre of the rod. Then we attached a 0.2 kg mass to the bottom of a string and wound the string around the vertical shaft of the apparatus, so that the bottom of the weight rose to the bottom edge of the tabletop the apparatus was on. We measured the distance from the bottom of the weight to the floor, and then let the weight fall to the floor, and measured the time it took to do so. We repeated this measurement, using the same initial height, four more times. Then we took the masses and the wingnuts off the horizontal rod and let the 0.2 kg mass fall in the same way as before, five times. Then we replaced the masses and the wingnuts, but put them both on the edges of the horizontal rod, and repeated the same falling mass measurements five times. We moved the masses in towards the centre of the rod and continued to repeat the falling mass measurements. We moved the masses in toward the centre four more times, each time taking five falling mass measurements. Data First Pass: m1 = 0.06158 kg; r1 = 0.162 m; m2 = 0.06158 kg; r2 = 0.115 m h = 0.876 m; R = 0.0064 m; M = 0.2 kg tavg(s) Δtavg(s) t0avg (s) Δt0avg(s) Im(energy transfer) (Js2) ΔIm (e.t.) (Js2) Im (point slope) ΔIm (p.m.)(Js2) (Js2) 8.436 0.34 4.236 0.106 0.0024 3.15 x 10-4 0.0024 3.41 x 10-6 -4 [(2x x 10-6/6.4 x 10-4) + (1 x 10ΔIm (e.t.) = Im [(2ΔR/R) + (Δh/h) + (2Δt/t) + (2Δt0/t0)] = 2.44 x 10 1 4/.0876) + (2x x 0.34/8.436) + (2x…
- 1135 Words
- 5 Pages
Good Essays