Ucm Ap Physic 1lab

AP Physic 1 Circular Motion

By Phatsakorn Rodphol Ms. Susana Alulod Mahidol University International Demonstration School Semester 1 Academic year 2018-2019

Objective To show how does the centripetal force (Fc) depend on tangential velocity (Vc) and the radius (R) and show how mass (m) affect circular motion.

Introduction The circular motion is the motion that the object is moving in the circular path. The circular motion can be divided into two type : uniform and non-uniform circular motion. First, the uniform circular motion is the motion of the object that in a circle at the constant speed (magnitude), but the velocity (direction) is changing. It can also be known the horizontal circular motion. At all position of the object in the uniform circular motion, the velocity will always tangent to the circle. This can also be call the tangential velocity. The changing in the direction of the velocity create the centripetal acceleration. Centripetal acceleration is the vector that point toward the center of the circle (center seeking) and perpendicular to the tangential velocity. Furthermore, the centripetal acceleration is the acceleration that change the direction of the object (make the object moving in the circular path). The formula of the centriple acceleration is a c =

v2 r

The centripletal acceleration is the thing that craete the centriplal force. The centripetal force is the net force that acts on the object to keep it moving along the circular path. According to the Newton Second Law’s: ΣF = ma . As the centripetal acceleration is a c = 2

centripetal force is ΣF = m( vr ) .

v2 r

. The

The non uniform circular motion is the motion that the centripetal velocity is keep changing along the circular motion because the object in this motion is under the influence of gravitational acceleration. It can also be called the vertical circular motion. The velocity will largest at the bottom of the motion and it will lowest at the top of the motion.

In this experiment, I use the centripetal apparatus to examine the different of centripetal force and centripetal acceleration when changing the mass and the radius of the circular motion.

Material 1. 100 grams Weights

2. Centripetal Force Apparatus

3. LabQuest

4. Stop Watch

Setting

Procedure 1. Set the radius to be 16 cm 2. Add 100 gram mass 3. Sustain the centripetal force of 1 newton 4. Measure the period 5. Add 100 gram mass 6. Follow step 3 and repeat it 7. Reduce the radius to 8 cm 8. Repeat step 2

Data & results Radius: 16 cm Mass (g) Period (T)

100

200

300

60 66

60 55

60 54

100

200

300

10 16

10 14

10 10

Radius: 8 cm Mass (g) Period (T)

Calculation

Radius 16 cm, mass 100 g

v=

2πr T

=

2π×16×10 −2 0.909

a=

v2 r

=

1.105 2 16×10 −2

Fc =

mv 2 r

=

= 1.105 m/s

= 7.63 m/ s2

100×10 −3 ×1.105 2 16×10 −2

= 0.76 N

Radius 16 cm, mass 200 g

v=

2πr T

=

2π×16×10 −2 1.09

a=

v2 r

=

0.92 2 16×10 −2

Fc =

mv 2 r

=

= 0.92 m/s

= 5.29 m/ s2

200×10 −3 ×0.92 2 16×10 −2

= 1.058 N

Radius 16 cm, mass 300 g

v=

2πr T

=

2π×16×10 −2 1.11

a=

v2 r

=

0.905 2 16×10 −2

Fc =

mv 2 r

=

= 0.905 m/s

= 5.11 m/ s2

300×10 −3 ×0.905 2 16×10 −2

= 1.535 N

Radius 8 cm, mass 100 g

v=

2πr T

=

2π×8×10 −2 0.625

a=

v2 r

=

0.804 2 8×10 −2

Fc =

mv 2 r

=

= 0.804 m/s

= 8.08 m/ s2

100×10 −3 ×0.804 2 8×10 −2

= 0.808 N

Radius 8 cm, mass 200 g

v=

2πr T

=

a=

v2 r

=

Fc =

mv 2 r

2π×8×10 −2 0.714 0.7039 2 8×10 −2

=

= 0.7039 m/s

= 6.19 m/ s2

200×10 −3 ×0.7039 2 8×10 −2

= 1.2387 N

Radius 8 cm, mass 300 g

v=

2πr T

=

2π×8×10 −2 1

a=

v2 r

=

0.5027 2 8×10 −2

Fc =

mv 2 r

=

= 0.5027 m/s

= 3.158 m/ s2

300×10 −3 ×0.5027 2 8×10 −2

= 0.947 N

Analysis As the mass increase, the centripetal force also increase according to the formula of 2

centripetal force ΣF = m( vr ) . The mass is directly proportional to the centripetal force. But, the acceleration is decreased ΣF = ma . The mass is inversely porpotional to the acceleration. As the mass is increase, the acceleration will be decrease. Furthermore, the velocity also decrease according to a c =

v2 r

. As the acceleration is decrease because of the

increase in mass, this make the decrease in velocity because the radius is constant. In the similar way, if we decrease the radius of the circular path, the centripetal force will increase. In this experiment, they are some of the error of the value on the calculate of radius 8 cm, mass 300g because the value of its centripetal force is lower than in the Radius 8 cm, mass 200 g. This error happen because we may swing Centripetal Force Apparatus not constantly (at the low speed).

Conclusion If the mass of the object in the uniform circular motion is increased, the centripetal force that point into the center of the circle also increase. Furthermore, if the radius of the circular path is decrease, the centripetal force will increase according to the formula of 2

centripetal force ΣF = m( vr ) . As the mass is increase, the centripetal accleration will decrease

Recommendation In this experiment, I suggest that it will be better to use the centripetal force apparatus that can swing the object with the constant speed. Therefore, this will lower the error in the calculation of centripetal force, centripetal acceleration, and tangential velocity.

Reference -

Uniform Circular Motion. (n.d.). Retrieved from https://www.physicsclassroom.com/mmedia/circmot/ucm.cfm

-

Centripetal force and gravitation | Physics | Science. (n.d.). Retrieved from https://www.khanacademy.org/science/physics/centripetal-force-and-gravitation/moda l/a/what-is-centripetal-force