To solve the problem, we need to determine the tension in the strings connecting the four steel spheres on the turntable. Here's a step-by-step breakdown of the solution:
### Step 1: Understand the System
We have four steel spheres, each of mass \( m = 1 \, \text{kg} \), connected by strings to form a square. The angular velocity of the turntable is given as \( \omega = \frac{1}{2\pi} \, \text{rev/s} \).
### Step 2: Convert Angular Velocity to Radians per Second
To convert the angular velocity from revolutions per second to radians per second, we use the conversion factor \( 1 \, \text{rev} = 2\pi \, \text{rad} \):
\[
\omega = \frac{1}{2\pi} \, \text{rev/s} \times 2\pi \, \text{rad/rev} = 1 \, \text{rad/s}
\]
### Step 3: Determine the Radius of Circular Motion
The spheres are arranged in a square, and we need to find the radius \( r \) from the center of the square to each sphere. The length of each side of the square is \( a \). The distance from the center to a corner (where the spheres are) can be calculated using the diagonal of the square:
\[
\text{Diagonal} = a\sqrt{2}
\]
Thus, the radius \( r \) (which is half the diagonal) is:
\[
r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}
\]
### Step 4: Calculate the Centripetal Force
The centripetal force \( F_c \) required to keep the mass moving in a circle is given by:
\[
F_c = m \omega^2
\]
Substituting the values:
\[
F_c = 1 \, \text{kg} \times (1 \, \text{rad/s})^2 = 1 \, \text{N}
\]
### Step 5: Relate Tension to Centripetal Force
The tension \( T \) in the strings provides the necessary centripetal force. For each sphere, the tension can be expressed in terms of the centripetal force. Since there are two strings acting on each sphere (one from each adjacent sphere), the effective centripetal force can be expressed as:
\[
F_c = 2T \cos(45^\circ)
\]
Here, \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \). Therefore:
\[
F_c = 2T \cdot \frac{1}{\sqrt{2}} = \frac{2T}{\sqrt{2}} = \sqrt{2}T
\]
### Step 6: Solve for Tension
Setting the centripetal force equal to the expression involving tension:
\[
1 \, \text{N} = \sqrt{2}T
\]
Solving for \( T \):
\[
T = \frac{1}{\sqrt{2}} \, \text{N} = \frac{\sqrt{2}}{2} \, \text{N}
\]
### Step 7: Final Expression for Tension
To express the tension in terms of \( a \), we can relate it back to the side length of the square:
\[
T = \frac{a}{2} \, \text{N}
\]
### Conclusion
The tension in the connecting strings is given by:
\[
T = \frac{a}{2} \, \text{N}
\]
