Figure 6-43 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 0.060 kg, the string has length L = 0.90 m and negligible mass, and the bob follows a circular path of circumference 0.90 m.
What is the period of the motion? The tension in the string is 0.596N.
The half-angle at the apex of the cone is theta = sin^-1[0.9/2 pi)/0.9]
= sin^-1(1/2 pi)
= 9.1 degrees
See http://en.wikipedia.org/wiki/Conical_pendulum
for the period of a conical pendulum. At this rather small cone angle, the period will be nearly the same as that of a normal pendualum of the same length.
4sin^2(theta) - 2sin(theta) - 2 = 0 on the interval 0 (less than or equal to ) theta < 2(pi)?
To find the period of the motion, we can use the formula for the period of a conical pendulum:
T = 2π√(L / g)
Where:
T = Period (in seconds)
L = Length of the string (in meters)
g = Acceleration due to gravity (approximately 9.8 m/s²)
In this case, we are given the length of the string (L = 0.90 m), and the acceleration due to gravity is a constant value (g = 9.8 m/s²).
So, let's substitute the given values into the formula and calculate the period:
T = 2π√(0.90 / 9.8)
T = 2π√0.0918
T = 2π * 0.303
T ≈ 1.91 seconds
Therefore, the period of the motion is approximately 1.91 seconds.