A small 0.469-kg object moves on a frictionless horizontal table in a circular path of radius 1.42 m. The angular speed is 6.31 rad/s. The object is attached to a string of negligible mass that passes through a small hole in the table at the center of the circle. Someone under the table begins to pull the string downward to make the circle smaller. If the string will tolerate a tension of no more than 105 N, what is the radius of the smallest possible circle on which the object can move?
Centripetal acceleration a=ω²•R=T/m
R=T/m•ω²
The law of conservation of angular momentum
I• ω²=I₀•ω₀²
ω=(I₀/I)•ω₀²=(R₀²/R²)•ω₀
R³=(m•R⁴•ω₀²)/T ,
R=...
To determine the smallest possible circle on which the object can move, we need to find the tension in the string when the circle is at its smallest.
First, let's calculate the initial tension in the string when the object is moving in a circle of radius 1.42 m.
The centripetal force required to keep the object moving in a circle is provided by the tension in the string, and it can be calculated using the formula:
Tension = (mass * velocity^2) / radius
Given:
Mass (m) = 0.469 kg
Angular speed (ω) = 6.31 rad/s
Radius (r) = 1.42 m
To find velocity (v), we can use the relationship between angular speed and linear speed: v = r * ω.
Substituting the values, we have:
v = 1.42 m * 6.31 rad/s
v ≈ 8.97 m/s
Now, we can calculate the initial tension (T_initial) using the formula:
T_initial = (mass * velocity^2) / radius
Substituting the values, we have:
T_initial = (0.469 kg * (8.97 m/s)^2) / 1.42 m
T_initial ≈ 2.972 N
Next, let's determine the tension in the string when the object is moving on the smallest possible circle. At the smallest circle, the centripetal force required to maintain the circular motion will be highest, and the tension in the string will also be at its highest.
We know that the maximum tension the string can tolerate is 105 N.
So, we need to find the maximum force the object can experience at the smallest possible circle. Since the force is provided by the tension in the string, we can use the formula:
Force = mass * acceleration
The acceleration can be calculated using the centripetal acceleration formula:
Acceleration = (velocity^2) / radius
We can rearrange the formula to find the maximum force:
Force = mass * ((velocity^2) / radius)
Substituting the values, we have:
Force = 0.469 kg * ((8.97 m/s)^2) / radius
We can set the maximum force equal to the tension in the string when the circle is at its smallest:
105 N = 0.469 kg * ((8.97 m/s)^2) / radius
Now, we can solve for the radius:
radius = 0.469 kg * ((8.97 m/s)^2) / 105 N
radius ≈ 0.035 m
Therefore, the smallest possible radius on which the object can move is approximately 0.035 m.