The Applications of Equations of Motion and Equations of Angular Motion - Lab Report Example

Introduction In linear motion the objects under consideration move in a straight line. The objects can either be moving with uniform velocity or may be under acceleration. A good example of linear motion with constant velocity is the case of a trolley on a frictionless plane will an example of uniform acceleration is a free fall a object in the air to the ground (Tipler P.A.& Mosca G. 2003). Experiences of circular motion are a common experience in a persons every day life. Some of the equipment in leisure parks such as “pirate ship” loop the loop and the wave swinger involve circular motion. Other common experiences of circular motion are in the cornering of a bicycle or a car, the function of washing machines, salad driers, governors of steam engines and tumble and spin driers (Knudsen, M.& Hjorth, G. ,2000). This experiment is an attempt to analyze circular motion so that to be able to model the situations experienced in circular motion and to be able to make prediction and give explanation to observation that can be made. It was considered to y philosophers of the ancient world that circular motion was a natural motion. They had the notion that all the heavenly bodies, the planets and all the stars were in circular motion around the earth (Sabol, et al 1999). It was believed that once the bodies were set in their paths by the Gods who were believed to be all powerful, the bodies would continue to move round with no need for further investigation. Newton however had a contrary point of view to that of the ancient philosophers, that circular motion required some force for sustenance. The first law on Newton’s First law points out that any body will tend to remain in a state of rest or would remain in a state of uniform motion unless there is external force acting on it. For the case involving a body in circular motion, the body is obviously not at rest and neither is it motion having a constant speed owing to the fact that for any body even when the speed is constant or not there is constant change it the direction of motion. From the Newton’s First Law, it is clear that there must be a resultant force that acts on a body which is quantified by Newton’s Second law as being the mass multiplied by the acceleration of the body in consideration. Theory for linear motion This experiment involves learning about kinematics of bodies travelling with uniform velocity or objects travelling with uniform acceleration. The most common example of uniform acceleration the case of free falling objects where the only force influencing the movement of the object is the force of gravity. The object under this circumstance will be accelerating due to the force of gravity g. near the earth source the force of gravity is relatively constant with a value g= 9.81m/s2 (Nainan K. ,2009). Air resistance may have some influence on the acceleration of free falling objects. However, the effect of air resistance is negligible where objects with high density are involved (Resnick, R. & Halliday, D.,1966). The equation that are dealt with under linear motion are i.  ii.  iii.  iv.  Where v is final velocity, u is the initial velocity, t is time taken, s is the distance covered and a is acceleration The objective of the experiment is to study the relationship governing position of an object, its velocity and acceleration for linear motion in the case where the objects are moving with constant velocity and the where objects are moving with constant acceleration. Theory behind circular motion Objects can be seen to be in uniform circular motion with their speed being constant (but their velocity not being constant) and also their acceleration being constant, a good example being the case where a mass is being whirled at the end of a string. As the mass attached to the end of the string is constantly changing direction, the velocity therefore is also constantly changing as the velocity is a vector quantity with both magnitude and direction and any in the two elements means change in velocity. There is a constant change of the direction of the mass towards the centre of the circle being described by the motion of the mass. This therefore means that the mass is experiencing a constant acceleration towards the centre of the circle. The acceleration towards the centre of the circle referred to as centripetal acceleration  is given by the expression Figure 1 The centripetal force F=ma and thus if the centripetal acceleration is known the centripetal force can also be calculated. Procedure for linear motion experiment Horizontal plane In the first case a cart was placed on a frictionless flat plane. The cart was then pushed slightly to make it move with a constant velocity. A watch was used to find the time and position of the cart on the plane. The Inclined plane The second case involves the cart moving on an inclined plane. In figure the cart is in motion on a frictionless inclined plane with its acceleration being constant. The acceleration of the cart is given by a=. The value of angle  is found by relating the height h and the length of the plane l. Thus  =. A stop watch was used in locating the position of the cart at different times. The height h = 13cm and l = 150cm Results Table 1: Results for the flat plane Distance (cm) Time in milliseconds 20 20 40 40 60 60 80 80 100 100 120 120 140 141 160 162 180 183 200 204 Time taken to cover the first 20cm (0.2m) = 20ms = 0.33s Velocity of cart for first 20cm =  Time taken to cover between 80cm and 100cm = 100-80 =20ms = 0.33s Velocity of cart between 80cm distance and 100cm distance =  The last 20 cm is from 180cm to point 200cm The time taken to cover the last 20cm = 205-183 = 22ms= 0.367 Velocity of cart for the last =  Acceleration between the 20cm distance and 100cm distance = 0 since difference in velocity at distance 0 to 20 and 80 to 100 is zero. Acceleration between 20cm distance and 200cm=  Table 2 Results for inclined plane Distance Time s 20 0.68 40 0.97 60 1.18 80 1.37 100 1.53 120 1.68 140 1.81 Velocity of cart between 20cm 40cm=  Time taken to cover between 120cm and 140cm = 1.81-1.68 =1ms = 0.13s Velocity of cart between 120cm mark and 140cm mark = Acceleration between 120cm mark and 140cm mark = Acceleration=  Circular motion Apparatus Variable speed turn table with circular lines at various radius from the centre of disc Five circular discs with a centre whole and ranging 5g to 25g in mass A tachometer Procedure The smallest mass was placed at a circle with the smallest radius on the disc. The disc was then set in motion with the speed being increased in small increments with a tachometer held at the top axis of the turntable being used measure speed of the turntable. The speed at which the mass is swept away from the table is noted. The second mass is placed mass was placed at the same position and the speed of release of the mass in noted. This procedure was done for the remaining masses with the speed being noted. After performing the procedure on the first radius on the turn table the same procedure was followed for the other radius on the turn table with the velocity of release for the various masses being noted. Force required for dislodging a mass disc from the turn tables The force required to dislodge a disc was found by a touching the disc to a sensitive spring balance using a small string. Glue was used to stick the string on the side of mass discs. The other end of the string was then tied to the sensitive spring balance. The spring was then pulled in a horizontal position while noting the reading on the scale at the point when the mass is first set in motion. This procedure was done for the five masses and the reading on the scale was noted. Results Table 3 Distance of mass from centre of disc = 10cm Mass rpm v 10 60.4 0.632187 3.9966 0.039966 20 59.5 0.622767 3.878383 0.077568 30 60.9 0.63742 4.063043 0.121891 40 60.4 0.632187 3.9966 0.159864 50 60.4 0.632187 3.9966 0.19983 Table 5 Distance of mass from centre of disc = 15cm Mass rpm v 10 74.0 0.774533 3.999346 0.039993 20 73.0 0.764067 3.891986 0.07784 30 74.5 0.779767 4.053574 0.121607 40 74.0 0.774533 3.999346 0.159974 50 74.0 0.774533 3.999346 0.199967 Table 6 Distance of mass from centre of disc = 20cm Mass rpm v 10 85.5 0.8949 4.00423 0.040042 20 84.2 0.881293 3.88339 0.077668 30 86.0 0.900133 4.0512 0.121536 40 85.5 0.8949 4.00423 0.160169 50 85.5 0.8949 4.00423 0.200212 Table 7 Distance of mass from centre of disc = 25cm Mass rpm v 10 95.5 0.999567 3.996534 0.039965 20 94.5 0.9891 3.913275 0.078266 30 96.5 1.010033 4.080669 0.12242 40 95.5 0.999567 3.996534 0.159861 50 95.5 0.999567 3.996534 0.199827 Table 8 Distance of mass from centre of disc = 30cm Mass rpm v 10 104.5 1.093767 3.987752 0.039878 20 103.1 1.079113 3.881619 0.077632 30 105.5 1.104233 4.064438 0.121933 40 104.5 1.093767 3.987752 0.15951 50 104.5 1.093767 3.987752 0.199388 Table 9 Force for dislodging disc Mass Force 1 Force 2 Force 3 Average force 10 0.042 0.044 0.04 0.042 20 0.086 0.078 0.082 0.082 30 0.12 0.12 0.124 0.121 40 0.166 0.154 0.16 0.16 50 0.192 0.198 0.196 0.195 Analysis and discussion In the case where the trolley was on a straight line it was seen that the results was close to the ideal situation where the motion was supposed to be frictionless. In the ideal frictionless situation the ideal time to cover the distances shown in column 1 of table 10 are as shown column 3. It can be seen clearly that up to a distance of 120 cm the ideal time was equal to the actual time recorded in the experiment. Table 10: Results for the flat plane Distance (cm) Time in ms (actual) Ideal 20 20 20 40 40 40 60 60 60 80 80 80 100 100 100 120 120 120 140 141 140 160 162 160 180 183 180 200 204 200 In the case of the inclined plane the g is calculated from the experimental results as  10.9 This value is very close to the actual value of g which is 9.81 This shows that the experiment was successful and thus the formula of acceleration along an inclined plane is validated. Comparison of circular motion results In the circular motion experiment it has been seen when the mass is placed near the centre of the turn table the rpm is generally lower compared to the case when the mass is away from the centre. For example for a mass of 10g placed 10cm from the centre the rpm= 60.4 while the same mass placed 30cm away from the centre rpm= 104.5. It is also seen from the results that the rpm do not vary with the heaviness of mass. The velocity v is higher when the distance of mass from the centre of turn table r is higher but the acceleration  remains constant. The centripetal force at the point of movement of the mass is proportional to the mass of the discs. The centripetal forces for the masses at various r and the forces required to manually dislodge the mass are as shown in table 11. From the table it can be see that the centripetal force does not much from the expected value which was found by trying to dislodge the masses manually. Table 11 mass Force at r=10 Force at r=15 Force at r=20 Force at r=25 Force at r=30 Manual dislodge 10g 0.039966 0.039993 0.040042 0.039965 0.039878 0.042 20g 0.077568 0.07784 0.077668 0.078266 0.077632 0.082 30g 0.121891 0.121607 0.121536 0.12242 0.121933 0.121 40g 0.159864 0.159974 0.160169 0.159861 0.15951 0.16 50g 0.19983 0.199967 0.200212 0.199827 0.199388 0.195 The slight variation in centripetal forces seen for the same mass at different r values could be attributed to some slight variation in friction between the wooden turntable disc and the mass at different points. It order to reduce this variation it is important to ensure that the surface of the turntable is uniform. Conclusion It can conclude from this experiment that it is possible to find the value of g experimentally using an inclined plane. From the circular motion experiment it is clear that centripetal forces are proportional to the mass of the object in circular motion. The experiments have successfully validated the theoretical relations in both linear and circular motion. References Knudsen, M.& Hjorth, G. (2000). Elements of Newtonian mechanics: including nonlinear dynamics (3 ed.). Springer. p. 96. ISBN 3-540-67652-X Nainan K. (2009).,Gravitation Resnick, R. & Halliday, D. (1966), Physics, Chapter 3 (Vol I and II, Combined edition), Wiley International Edition, Library of Congress Catalog Card No. 66-11527 Tipler P.A.& Mosca G. (2003). "Physics for Scientists and Engineers", Chapter 2 (5th edition), W. H. Freeman and company: New York and Basing stoke, Nainan K. (2008).Varghese, Hypothesis on MATTER (second edition). Bibliography Henry Semat(1958).: Circular Motion and Gravitation University of Nebraska – Lincoln Hill, G. W. (1878), “Researches in the Lunar Theory,” American Journal of Mathematics, Vol. 1, No. 1, , pp. 5–26, 129–147, 245–260. Howard IP, Heckmann T(1989);. Circular vection as a function of the relative sizes, distances, and positions of 2 competing visual displays. Perception 18(5):657–65. Rutherford F. J. (1981). Project Physics. New York University Sabol, C., et al (1999) “Satellite Formation Flying Design and Evolution,” Advances in the Astronautical Sciences, Vol. 102, No. 1, pp. 265–284. Nagpal G.R. (1997) Machine design. Khanna Publishers. Delhi. Read More

The acceleration of the cart is given by a=. The value of angle  is found by relating the height h and the length of the plane l. Thus  =. A stop watch was used in locating the position of the cart at different times. The height h = 13cm and l = 150cm Results Table 1: Results for the flat plane Distance (cm) Time in milliseconds 20 20 40 40 60 60 80 80 100 100 120 120 140 141 160 162 180 183 200 204 Time taken to cover the first 20cm (0.2m) = 20ms = 0.33s Velocity of cart for first 20cm =  Time taken to cover between 80cm and 100cm = 100-80 =20ms = 0.

33s Velocity of cart between 80cm distance and 100cm distance =  The last 20 cm is from 180cm to point 200cm The time taken to cover the last 20cm = 205-183 = 22ms= 0.367 Velocity of cart for the last =  Acceleration between the 20cm distance and 100cm distance = 0 since difference in velocity at distance 0 to 20 and 80 to 100 is zero. Acceleration between 20cm distance and 200cm=  Table 2 Results for inclined plane Distance Time s 20 0.68 40 0.97 60 1.18 80 1.37 100 1.53 120 1.68 140 1.

81 Velocity of cart between 20cm 40cm=  Time taken to cover between 120cm and 140cm = 1.81-1.68 =1ms = 0.13s Velocity of cart between 120cm mark and 140cm mark = Acceleration between 120cm mark and 140cm mark = Acceleration=  Circular motion Apparatus Variable speed turn table with circular lines at various radius from the centre of disc Five circular discs with a centre whole and ranging 5g to 25g in mass A tachometer Procedure The smallest mass was placed at a circle with the smallest radius on the disc.

The disc was then set in motion with the speed being increased in small increments with a tachometer held at the top axis of the turntable being used measure speed of the turntable. The speed at which the mass is swept away from the table is noted. The second mass is placed mass was placed at the same position and the speed of release of the mass in noted. This procedure was done for the remaining masses with the speed being noted. After performing the procedure on the first radius on the turn table the same procedure was followed for the other radius on the turn table with the velocity of release for the various masses being noted.

Force required for dislodging a mass disc from the turn tables The force required to dislodge a disc was found by a touching the disc to a sensitive spring balance using a small string. Glue was used to stick the string on the side of mass discs. The other end of the string was then tied to the sensitive spring balance. The spring was then pulled in a horizontal position while noting the reading on the scale at the point when the mass is first set in motion. This procedure was done for the five masses and the reading on the scale was noted.

Results Table 3 Distance of mass from centre of disc = 10cm Mass rpm v 10 60.4 0.632187 3.9966 0.039966 20 59.5 0.622767 3.878383 0.077568 30 60.9 0.63742 4.063043 0.121891 40 60.4 0.632187 3.9966 0.159864 50 60.4 0.632187 3.9966 0.19983 Table 5 Distance of mass from centre of disc = 15cm Mass rpm v 10 74.0 0.774533 3.999346 0.039993 20 73.0 0.764067 3.891986 0.07784 30 74.5 0.779767 4.053574 0.121607 40 74.0 0.774533 3.999346 0.159974 50 74.0 0.774533 3.999346 0.199967 Table 6 Distance of mass from centre of disc = 20cm Mass rpm v 10 85.5 0.8949 4.00423 0.040042 20 84.2 0.881293 3.88339 0.077668 30 86.0 0.900133 4.0512 0.121536 40 85.5 0.8949 4.00423 0.160169 50 85.5 0.8949 4.00423 0.200212 Table 7 Distance of mass from centre of disc = 25cm Mass rpm v 10 95.5 0.999567 3.996534 0.039965 20 94.5 0.9891 3.913275 0.078266 30 96.5 1.010033 4.080669 0.12242 40 95.5 0.999567 3.996534 0.159861 50 95.5 0.999567 3.996534 0.199827 Table 8 Distance of mass from centre of disc = 30cm Mass rpm v 10 104.5 1.093767 3.987752 0.039878 20 103.1 1.079113 3.881619 0.077632 30 105.5 1.104233 4.064438 0.121933 40 104.5 1.093767 3.987752 0.15951 50 104.5 1.093767 3.987752 0.199388 Table 9 Force for dislodging disc Mass Force 1 Force 2 Force 3 Average force 10 0.042 0.044 0.04 0.042 20 0.086 0.078 0.082 0.082 30 0.12 0.12 0.124 0.121 40 0.166 0.154 0.16 0.16 50 0.192 0.198 0.196 0.

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