What is the tension in a 1-meter string that is spinning a 0.5-kg rock in a horizontal circle 3 times per second?
To find the tension in the string, we need to consider the centripetal force acting on the rock. The centripetal force is provided by the tension in the string and is given by the equation:
F = (m * v^2) / r
where:
F is the centripetal force,
m is the mass of the object (rock),
v is the linear velocity of the object, and
r is the radius of the circular path.
First, we need to find the linear velocity of the rock. Since the rock completes 3 revolutions (cycles) per second, we can calculate the angular velocity (ω) as:
ω = 2π * f
where:
ω is the angular velocity,
π is a mathematical constant (approximately 3.14159), and
f is the frequency (number of revolutions per second).
Substituting the given frequency, we have:
ω = 2π * 3 = 6π rad/s
Next, we can calculate the linear velocity (v) using the relationship between linear and angular velocity:
v = r * ω
Here, the radius (r) is given as 1 meter:
v = 1 * 6π = 6π m/s
Now that we have the mass (m = 0.5 kg) and linear velocity (v = 6π m/s), we can calculate the centripetal force (F) using the centripetal force equation mentioned earlier:
F = (m * v^2) / r
F = (0.5kg * (6π m/s)^2) / 1m
Simplifying the expression:
F = (0.5kg * 36π^2 m^2/s^2) / 1m
F = 18π^2 N
The tension in the string is approximately 18π^2 Newtons.