A small 0.499 kg object moves on a frictionless horizontal table in a circular path of radius 0.96 m. The angular speed is 6.37 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 on which the object can move, we first need to determine the maximum tension force that the string can tolerate.

The tension force in the string provides the centripetal force needed to keep the object moving in a circular path. This force can be calculated using the equation:

T = m * a

where T is the tension force, m is the mass of the object, and a is the centripetal acceleration.

The centripetal acceleration can be calculated using the equation:

a = (v^2) / r

where v is the linear velocity of the object and r is the radius of the circular path.

We can find the linear velocity of the object using the equation:

v = ω * r

where ω is the angular speed of the object.

Substituting the values into the equations, we get:

v = (6.37 rad/s) * (0.96 m) = 6.11 m/s

a = (6.11 m/s)^2 / (0.96 m) = 39.09 m/s^2

Now, we can calculate the maximum tension force that the string can tolerate:

T = (0.499 kg) * (39.09 m/s^2) = 19.46 N

Since the string can tolerate a maximum tension force of 105 N, we can equate the maximum tension force with the centripetal force:

105 N = m * a

Solving for m:

m = (105 N) / (39.09 m/s^2) ≈ 2.685 kg

Now, we can find the radius of the smallest possible circle:

T = m * a

105 N = (2.685 kg) * (v^2 / r)

Rearranging the equation, we get:

r = (2.685 kg) * (v^2 / 105 N)

Substituting the values, we get:

r = (2.685 kg) * ((6.11 m/s)^2 / 105 N)

r ≈ 0.293 m

Therefore, the radius of the smallest possible circle on which the object can move is approximately 0.293 m.