Assumed mass of bob: m = 0.0640 kg Assumed length of pendulum L: = 6.4 m Trial 1 2 3 4 5 6 6 7 8 Radius L cos e R Trial 1 2 3 4 5 6 6 7 8 (m) (°) (m) 3.0 27.94 5.65 3.3 31.02 5.48 3.5 33.13 5.36 3.8 36.4 5.15 4.0 4.2 4.5 5.0 5.5 38.66 4.99 40.99 4.83 44.65 4.55 51.34 3.99 59.2 3.28 Height h (m) Expected period using Eq. (6) (s) 5.65 5.48 5.36 5.15 4.99 4.83 4.55 3.99 3.28 Angular frequency 2πt W= 4.77 4.70 4.65 4.55 4.48 4.41 4.28 4.01 3.63 T (/s) 1.32 1.34 1.35 1.38 1.40 1.42 1.47 1.57 1.73 Total time for Table 2. Centripetal Force and Tension in String 5 cycles (s) 23.88 23.61 23.37 22.85 22.47 22.12 21.53 20.05 18.16 Speed of bob v = Rw (m/s) Expt'l period T (s) 3.96 4.42 4.73 5.24 5.6 5.96 6.62 7.85 9.52 4.77 4.72 4.67 4.57 4.49 4.42 4.31 4.01 3.63 Fc = R (N) Square of expt'l Centripeta I force mv² 0.33 0.38 0.41 0.46 0.50 period 7² (S²) 0.54 0.62 0.79 1.05 22.75 22.28 21.81 20.88 20.16 19.54 18.58 16.08 13.18 % error of T Tension in string F sin = (N) 0.71 (%) 0 0.43 0.43 0.44 0.22 0.23 0.70 0 0 mv2 R
Assumed mass of bob: m = 0.0640 kg Assumed length of pendulum L: = 6.4 m Trial 1 2 3 4 5 6 6 7 8 Radius L cos e R Trial 1 2 3 4 5 6 6 7 8 (m) (°) (m) 3.0 27.94 5.65 3.3 31.02 5.48 3.5 33.13 5.36 3.8 36.4 5.15 4.0 4.2 4.5 5.0 5.5 38.66 4.99 40.99 4.83 44.65 4.55 51.34 3.99 59.2 3.28 Height h (m) Expected period using Eq. (6) (s) 5.65 5.48 5.36 5.15 4.99 4.83 4.55 3.99 3.28 Angular frequency 2πt W= 4.77 4.70 4.65 4.55 4.48 4.41 4.28 4.01 3.63 T (/s) 1.32 1.34 1.35 1.38 1.40 1.42 1.47 1.57 1.73 Total time for Table 2. Centripetal Force and Tension in String 5 cycles (s) 23.88 23.61 23.37 22.85 22.47 22.12 21.53 20.05 18.16 Speed of bob v = Rw (m/s) Expt'l period T (s) 3.96 4.42 4.73 5.24 5.6 5.96 6.62 7.85 9.52 4.77 4.72 4.67 4.57 4.49 4.42 4.31 4.01 3.63 Fc = R (N) Square of expt'l Centripeta I force mv² 0.33 0.38 0.41 0.46 0.50 period 7² (S²) 0.54 0.62 0.79 1.05 22.75 22.28 21.81 20.88 20.16 19.54 18.58 16.08 13.18 % error of T Tension in string F sin = (N) 0.71 (%) 0 0.43 0.43 0.44 0.22 0.23 0.70 0 0 mv2 R
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How do I solve for this? Number 1 is already given
Transcribed Image Text:Assumed mass of bob: m = 0.0640 kg
Assumed length of pendulum L: = 6.4 m
Square of
expt'l
e L cos e
Expt'l
time for period
Radius
Total
Expected
period
using Eq. (6)
(s)
%
R
error
Trial
of T
period 72
(s?)
(*)
(m)
cycles
(s)
(m)
(s)
(%)
1
3.0
27.94
5.65
4.77
23.88
4.77
22.75
2
3.3
31.02
5.48
4.70
23.61
4.72
22.28
0.43
3
3.5
33.13
5.36
4.65
23.37
4.67
21.81
0.43
4
3.8
36.4
5.15
4.55
22.85
4.57
20.88
0.44
4.0
38.66
4.99
4.48
22.47
4.49
20.16
0.22
6
4.2
40.99
4.83
4.41
22.12
4.42
19.54
0.23
4.5
44.65
4.55
4.28
21.53
4.31
18.58
0.70
5.0
51.34
3.99
4.01
20.05
4.01
16.08
8
5.5
59.2
3.28
3.63
18.16
3.63
13.18
Table 2. Centripetal Force and Tension in String
Angular
frequency
Speed of
bob
v = Rw
Centripeta
I force
Height
Tension in
string
Trial
h
mv2
F sin e =
T
Fc= R
(m)
(/s)
(m/s)
(N)
R
(N)
0.71
1
5.65
1.32
3.96
0.33
2
5.48
1.34
4.42
0.38
3
5.36
1.35
4.73
0.41
4
5.15
1.38
5.24
0.46
4.99
1.40
5.6
0.50
6
4.83
1.42
5.96
0.54
4.55
1.47
6.62
0.62
7
3.99
1.57
7.85
0.79
8
3.28
1.73
9.52
1.05
6700
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