A small 0.500 kg object moves on a frictionless horizontal table in a circular path of radius 1.00 m. The angular speed is `6.28 "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 toler ate a tension of no more than 105 N, what is the radius of the smallest possible circle on which the object can move?

Options

A. 0.376 m
B. 0.434 m
C. 0.573 m
D. 0.188 m

Answer

To solve the problem, we need to determine the radius of the smallest possible circle on which the object can move while ensuring that the tension in the string does not exceed 105 N. We will use the concepts of circular motion and centripetal force.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the object, \( m = 0.500 \, \text{kg} \)
- Angular speed, \( \omega = 6.28 \, \text{rad/s} \)
- Maximum tension in the string, \( T_{\text{max}} = 105 \, \text{N} \)

2. **Understand the relationship between tension and centripetal force:**
The tension in the string provides the necessary centripetal force to keep the object moving in a circular path. The centripetal force \( F_c \) required for circular motion is given by:
\[ F_c = m \cdot a_c \]
where \( a_c \) is the centripetal acceleration. The centripetal acceleration can also be expressed in terms of angular speed and radius \( r \):
\[ a_c = \omega^2 \cdot r \]
Therefore, we can write:
\[ F_c = m \cdot (\omega^2 \cdot r) \]

3. **Set the centripetal force equal to the tension:**
Since the tension is providing the centripetal force, we have:
\[ T = m \cdot (\omega^2 \cdot r) \]
Setting \( T \) equal to the maximum tension:
\[ 105 = 0.500 \cdot (6.28^2 \cdot r) \]

4. **Calculate \( \omega^2 \):**
First, calculate \( \omega^2 \):
\[ \omega^2 = (6.28)^2 = 39.4784 \, \text{rad}^2/\text{s}^2 \]

5. **Substitute \( \omega^2 \) into the tension equation:**
Substitute \( \omega^2 \) into the equation:
\[ 105 = 0.500 \cdot (39.4784 \cdot r) \]

6. **Solve for \( r \):**
Rearranging the equation to solve for \( r \):
\[ r = \frac{105}{0.500 \cdot 39.4784} \]
\[ r = \frac{105}{19.7392} \approx 5.31 \, \text{m} \]

### Final Answer:
The radius of the smallest possible circle on which the object can move is approximately **5.31 m**.