A 0.013 kg rubber stopper attached to a 0.93 m string is swung in a circle. If the tension in the string is 0.70 N, what is the period of the stopper's revolution?
Ac = v^2/r
so
.7 = .013 v^2/.93
solve for v
we know r
time = distance / speed = 2 pi r/v
Why did the rubber stopper join the circus? It wanted to be a "spin"-tacular performer!
Now, let's get down to the equation-taming business. We can use the formula for centripetal force to find the period of the stopper's revolution. The centripetal force is given by the equation F = m * (v^2 / r), where F is the tension in the string, m is the mass of the stopper, v is the velocity, and r is the radius of the circle.
In this case, the tension in the string is 0.70 N, the mass of the stopper is 0.013 kg, and the radius of the circle is 0.93 m.
We can rearrange the formula to solve for the velocity v: v = sqrt((F * r) / m).
Now we can plug in the values: v = sqrt((0.70 N * 0.93 m) / 0.013 kg).
Calculating this, we find that v ≈ 7.56 m/s.
The period of the stopper's revolution can be calculated using the formula T = (2 * π * r) / v.
Plugging in the values, T ≈ (2 * π * 0.93 m) / 7.56 m/s.
Calculating this, we find that T ≈ 0.78 seconds.
So, the period of the stopper's revolution is approximately 0.78 seconds. Happy spinning!
To find the period of the stopper's revolution, we can use the centripetal force formula. The centripetal force is provided by the tension in the string, and it is equal to the gravitational force acting on the stopper.
1. Calculate the gravitational force acting on the stopper:
F_gravity = mass * gravitational acceleration
= 0.013 kg * 9.8 m/s^2
≈ 0.1274 N
2. Set the centripetal force equal to the gravitational force:
F_centripetal = F_gravity
Tension = mass * (velocity^2 / radius)
0.70 N = 0.013 kg * (velocity^2 / 0.93 m)
3. Rearrange the equation to solve for velocity:
velocity^2 = (0.70 N * 0.93 m) / 0.013 kg
velocity^2 ≈ 49.3385 m^2/s^2
4. Take the square root of both sides to solve for velocity:
velocity ≈ √(49.3385 m^2/s^2)
velocity ≈ 7.02 m/s
5. The period of the stopper's revolution is given by:
period = (2π * radius) / velocity
= (2π * 0.93 m) / 7.02 m/s
≈ 0.838 seconds
Therefore, the period of the stopper's revolution is approximately 0.838 seconds.
To find the period of the stopper's revolution, we can use the equation for the period of a circular motion:
T = 2π√(r/g)
Where:
T is the period (time for one revolution)
π is the mathematical constant pi (approximately 3.14159)
r is the radius of the circular motion
g is the acceleration due to gravity (approximately 9.8 m/s^2)
In this problem, the radius of the circular motion is the length of the string, which is given as 0.93 m.
To continue, we need to find the value of acceleration due to gravity (g) in order to substitute it into the equation.
Acceleration due to gravity is the rate at which an object falls toward the Earth. Its approximate value is 9.8 m/s^2. We can use this value for most problems on Earth.
Next, we can substitute the values into the equation to find the period:
T = 2π√(0.93/9.8)
T = 2π√(0.0949)
T ≈ 2π * 0.308
T ≈ 1.93 seconds.
Therefore, the period of the stopper's revolution is approximately 1.93 seconds.