Asked by greg
Fig. 6-48 shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at constant speed. (The cord sweeps out a cone as the bob rotates.) The bob has a mass of 0.038 kg, the string has length L = 1.2 m and negligible mass, and the bob follows a circular path of circumference 0.75 m. What are (a) the tension in the string and (b) the period of the motion?
Answers
Answered by
drwls
The angle of the pendulum from vertical is
A = arcsin r/L = arcsin [0.75/(2 pi L)]
= 5.71 degrees
You will need this angle later.
The vertical and horizontanl equations of motion are:
T sin A = m V^2/R
T cos A = m g
which tells you that
tan A = 0.100 = V^2/gR
V^2 = 0.100*9.8m/s^2*0.1194m = 0.1170 m^2/s^2
V = 0.342 m/s
(a) T = mg/cosA = 0.038kg*9.8m/s^2/.9950
= 0.3743 Newtons
(b) Period = (circumference)/V
= 2 pi R/sqrt(g R tan A)
for small angles A, tan A = R/L, so
Period = 2 pi R /sqrt (g R *R/L)
= 2 pi sqrt (L/g)
which is the same as the formula for a pendulum oscillating in a plane (one dimension)
Period = 2.2 seconds
Check my work
A = arcsin r/L = arcsin [0.75/(2 pi L)]
= 5.71 degrees
You will need this angle later.
The vertical and horizontanl equations of motion are:
T sin A = m V^2/R
T cos A = m g
which tells you that
tan A = 0.100 = V^2/gR
V^2 = 0.100*9.8m/s^2*0.1194m = 0.1170 m^2/s^2
V = 0.342 m/s
(a) T = mg/cosA = 0.038kg*9.8m/s^2/.9950
= 0.3743 Newtons
(b) Period = (circumference)/V
= 2 pi R/sqrt(g R tan A)
for small angles A, tan A = R/L, so
Period = 2 pi R /sqrt (g R *R/L)
= 2 pi sqrt (L/g)
which is the same as the formula for a pendulum oscillating in a plane (one dimension)
Period = 2.2 seconds
Check my work
Answered by
greg
Thanks!
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