Centripetal Force
Title: Centripetal Force
Tools and Equipments: nylon cord, different weighing hanging masses, stopwatch, meter stick.
Purpose: To be able to determine the relationship between centripetal force, mass, velocity, and the radius of orbit for a body that is undergoing centripetal acceleration. To investigate the dynamics of uniform circular motion. Specifically the relationships among the centripetal force, the accelerated mass and the radius of rotation. Procedure:
THEORY:
DATA AND CALCULATION:
PART A (Constant radius) r = .20 m
Frequency = (30 Revolutions)/〖t 〗_(av (30 revolution)) → f = 30/(26.27 s) = 1.142 Hz
V= 2πrf = 2π (.20)(1.142) = 1.44 m/s
F = 〖mv〗^2/r = ((.456)(1.44)²)/(.20) = 4.73 N
Radius
(m) Mass
(kg) F_exp
(N) 〖Time〗_1
(s) 〖Time〗_2
(s) 〖Time〗_av
(s) Frequency
(Hz) Velocity
V=2πrf
(m/s) F_Th
F=〖mv〗^2/r (N)
.20 m .456 kg 4.46 N 26.21 s 26.33 s 26.27 s 1.142 Hz 1.44 m/s 4.73 N
.20 m .506 kg 4.46 N 27.65 s 27.71 s 27.68 s 1.084 Hz 1.36 m/s 4.68 N
.20 m .556 kg 4.46 N 28.68 s 28.65 s 28.67 s 1.046 Hz 1.31 m/s 4.77 N
.20 m .606 kg 4.46 N 29.59 s 29.98 s 29.78 s 1.007 Hz 1.27 m/s 4.88 N
% Error F_Th vs F_exp
% Error = (F_(Th-) F_exp)/F_TH × 100 → = (4.73-4.46)/4.73 ×100 → = 5.71 %
F_TH F_exp % Error
4.73 N 4.46 N 5.71 %
4.68 N 4.46 N 4.70 %
4.77 N 4.46 N 6.50 %
4.88 N 4.46 N 8.61 %
PART B (Constant mass) m = .456 kg
Mass
(kg) Radius
(m) F_exp (N) 〖Time〗_1
(s) 〖Time〗_2
(s) 〖Time〗_av
(s) Frequency
(Hz) Velocity
V=2πrf
(m/s) F_Th
F=〖mv〗^2/r
(N)
.456 kg .20 m 4.46 N 26.21 s 26.33 s 26.27 s 1.142 Hz 1.44 m/s 4.73 N
.456 kg .18 m 4.12 N 26.48 s 26.65 s 26.57 s 1.129 Hz 1.28 m/s 4.15 N
.456 kg .16 m 3.43 N 28.26 s 28.38 s 28.32 s 1.059 Hz 1.06 m/s 3.20 N
.456 kg .14 m 2.06 N 33.46 s 32.46 s 32.96 s .9102 Hz .801 m/s 2.09 N
% Error F_Th
Tools and Equipments: nylon cord, different weighing hanging masses, stopwatch, meter stick.
Purpose: To be able to determine the relationship between centripetal force, mass, velocity, and the radius of orbit for a body that is undergoing centripetal acceleration. To investigate the dynamics of uniform circular motion. Specifically the relationships among the centripetal force, the accelerated mass and the radius of rotation. Procedure:
THEORY:
DATA AND CALCULATION:
PART A (Constant radius) r = .20 m
Frequency = (30 Revolutions)/〖t 〗_(av (30 revolution)) → f = 30/(26.27 s) = 1.142 Hz
V= 2πrf = 2π (.20)(1.142) = 1.44 m/s
F = 〖mv〗^2/r = ((.456)(1.44)²)/(.20) = 4.73 N
Radius
(m) Mass
(kg) F_exp
(N) 〖Time〗_1
(s) 〖Time〗_2
(s) 〖Time〗_av
(s) Frequency
(Hz) Velocity
V=2πrf
(m/s) F_Th
F=〖mv〗^2/r (N)
.20 m .456 kg 4.46 N 26.21 s 26.33 s 26.27 s 1.142 Hz 1.44 m/s 4.73 N
.20 m .506 kg 4.46 N 27.65 s 27.71 s 27.68 s 1.084 Hz 1.36 m/s 4.68 N
.20 m .556 kg 4.46 N 28.68 s 28.65 s 28.67 s 1.046 Hz 1.31 m/s 4.77 N
.20 m .606 kg 4.46 N 29.59 s 29.98 s 29.78 s 1.007 Hz 1.27 m/s 4.88 N
% Error F_Th vs F_exp
% Error = (F_(Th-) F_exp)/F_TH × 100 → = (4.73-4.46)/4.73 ×100 → = 5.71 %
F_TH F_exp % Error
4.73 N 4.46 N 5.71 %
4.68 N 4.46 N 4.70 %
4.77 N 4.46 N 6.50 %
4.88 N 4.46 N 8.61 %
PART B (Constant mass) m = .456 kg
Mass
(kg) Radius
(m) F_exp (N) 〖Time〗_1
(s) 〖Time〗_2
(s) 〖Time〗_av
(s) Frequency
(Hz) Velocity
V=2πrf
(m/s) F_Th
F=〖mv〗^2/r
(N)
.456 kg .20 m 4.46 N 26.21 s 26.33 s 26.27 s 1.142 Hz 1.44 m/s 4.73 N
.456 kg .18 m 4.12 N 26.48 s 26.65 s 26.57 s 1.129 Hz 1.28 m/s 4.15 N
.456 kg .16 m 3.43 N 28.26 s 28.38 s 28.32 s 1.059 Hz 1.06 m/s 3.20 N
.456 kg .14 m 2.06 N 33.46 s 32.46 s 32.96 s .9102 Hz .801 m/s 2.09 N
% Error F_Th