Posted by greg on Saturday, March 1, 2008 at 11:26pm.
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?
- PHYSICS - drwls, Sunday, March 2, 2008 at 1:00am
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
- PHYSICS - greg, Sunday, March 2, 2008 at 1:26am
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