To solve the problem step by step, we will analyze the forces acting on the two masses and apply the concepts of circular motion and equilibrium.
### Step 1: Identify the given values
- Mass \( m_1 = 70 \, \text{g} = 0.07 \, \text{kg} \)
- Mass \( m_2 = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Total length of the string \( L = 1 \, \text{m} \)
- Length of the hanging part \( L_h = 60 \, \text{cm} = 0.6 \, \text{m} \)
### Step 2: Calculate the radius of the circular motion
The length of the string that rotates in a circle is given by:
\[
r = L - L_h = 1 \, \text{m} - 0.6 \, \text{m} = 0.4 \, \text{m}
\]
### Step 3: Analyze the forces acting on the hanging mass \( m_2 \)
The forces acting on mass \( m_2 \) are:
- Weight \( W = m_2 g = 0.1 \, \text{kg} \times 10 \, \text{m/s}^2 = 1 \, \text{N} \)
- Tension \( T \) in the string
For the system to be in equilibrium:
\[
T = m_2 g = 1 \, \text{N}
\]
### Step 4: Analyze the forces acting on the rotating mass \( m_1 \)
For mass \( m_1 \) which is rotating in a circle, the tension \( T \) also provides the centripetal force required for circular motion:
\[
T = \frac{m_1 v^2}{r}
\]
Substituting \( T = 1 \, \text{N} \):
\[
1 = \frac{0.07 \, v^2}{0.4}
\]
### Step 5: Solve for the velocity \( v \)
Rearranging the equation gives:
\[
v^2 = \frac{1 \times 0.4}{0.07}
\]
\[
v^2 = \frac{0.4}{0.07} \approx 5.7143
\]
\[
v = \sqrt{5.7143} \approx 2.39 \, \text{m/s}
\]
### Step 6: Relate linear velocity to angular frequency
The relationship between linear velocity \( v \) and angular frequency \( \omega \) is given by:
\[
v = r \omega \implies \omega = \frac{v}{r}
\]
Substituting the values:
\[
\omega = \frac{2.39}{0.4} \approx 5.975 \, \text{rad/s}
\]
### Step 7: Calculate the frequency \( f \)
The frequency \( f \) is related to angular frequency \( \omega \) by:
\[
f = \frac{\omega}{2\pi}
\]
Substituting the value of \( \omega \):
\[
f = \frac{5.975}{2\pi} \approx \frac{5.975}{6.2832} \approx 0.951 \, \text{Hz}
\]
### Final Answer
The frequency of rotation of the smaller mass \( m_1 \) for the system to be in equilibrium is approximately \( 0.951 \, \text{Hz} \).
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