A small 0.50 kg object moves on a frictionless horizontal table in a circular path of radius 1.0 m. the angular speed is 5.8 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 110. N what is the radius of the smallest possible circle on which the object can move?

Question

A small 0.50 kg object moves on a frictionless horizontal table in a circular path of radius 1.0 m. the angular speed is 5.8 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 110. N what is the radius of the smallest possible circle on which the object can move?

I have attempted this for about 45 minutes and came up with nothing because you do not get final angular velocity.

Answer 1

Sounds like conservation of angular momentum:

.5 * 1 * 5.8 = .5 r^2 omegaf
r^2 omegaf = 5.8
And forces
100 = .5 r omegaf^2
r omegaf^2 = 220

Two eq two unknowns (I get r = .52 omegaf = 20.3)

Answer 2

To solve this problem, we need to apply the principles of circular motion and equilibrium. We know that the object is moving in a circular path and that the force causing this circular motion is provided by the tension in the string.

Let's start by writing down the known information:
- Mass of the object (m) = 0.50 kg
- Angular speed (ω) = 5.8 rad/s
- Maximum tension in the string (T_max) = 110 N

Now, let's analyze the forces acting on the object at the smallest possible circle:

1. Centripetal force:
The tension in the string is the centripetal force acting on the object, given by:
Tension (T) = m * (ω^2) * r
where r is the radius of the circle.

2. Maximum tension:
We also know that the tension in the string should not exceed the maximum tolerance value:
T_max = 110 N

Now, we can set up an equation using the above information and solve it for the radius (r):

T = m * (ω^2) * r

Plugging in the known values:
110 N = 0.50 kg * (5.8 rad/s)^2 * r

Simplifying the equation:
110 N = 0.5 kg * 33.64 rad^2/s^2 * r

Dividing both sides by 0.5 kg * 33.64 rad^2/s^2:
r = 110 N / (0.5 kg * 33.64 rad^2/s^2)

Now, we can calculate the value of r:

r ≈ 3.272 meters (rounded to three decimal places)

Therefore, the radius of the smallest possible circle on which the object can move is approximately 3.272 meters.