A small 0.482-kg object moves on a frictionless horizontal table in a circular path of radius 0.77 m. The angular speed is 6.18 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?
To find the radius of the smallest possible circle, we need to determine the tension in the string when the circle becomes smallest. We can start by analyzing the forces acting on the object.
The only force acting on the object is the tension in the string since the table is assumed to be frictionless and no other forces are mentioned. The centripetal force that keeps the object moving in a circular path is provided by the tension in the string.
The centripetal force (Fc) is given by the formula:
Fc = m * ω^2 * r
Where:
m = mass of the object (0.482 kg)
ω = angular speed (6.18 rad/s)
r = radius of the circular path
Now, we need to find the maximum tension that the string can tolerate (Tmax), which is given as 105 N.
Since the object is attached to the string and the person is pulling downward, we can assume that the tension force (T) is directed radially inward. Thus, when the circle becomes smallest, the tension force is at its maximum.
Now we can set up an equation equating the centripetal force to the maximum tension:
Fc = Tmax
Substituting the formula for centripetal force:
m * ω^2 * r = Tmax
Plugging in the values:
(0.482 kg) * (6.18 rad/s)^2 * r = 105 N
Now, we can solve for the radius (r):
r = 105 N / [(0.482 kg) * (6.18 rad/s)^2]
Calculating this expression gives us the radius of the smallest possible circle on which the object can move.