A chain of 125 links is 1.25 m long and has mass of 2 kg with the ends fastened together. It is set for rotating at 50 revolutions per second. The centripetal force on each link is

Options

A. 3.14 N
B. 0.314 N
C. 314 N
D. None of these

Answer

To find the centripetal force on each link of the chain, we can follow these steps:

### Step 1: Understand the Problem
We have a chain of 125 links that is 1.25 m long and has a mass of 2 kg. The chain is rotating at a frequency of 50 revolutions per second. We need to find the centripetal force acting on each link.

### Step 2: Calculate the Radius of the Circle
When the chain is fastened together, it forms a circle. The circumference of the circle (C) is equal to the length of the chain (L):
\[ C = L = 1.25 \, \text{m} \]
The circumference of a circle is given by:
\[ C = 2\pi r \]
We can set these equal to find the radius \( r \):
\[ 1.25 = 2\pi r \]
\[ r = \frac{1.25}{2\pi} \]

### Step 3: Calculate the Angular Velocity (ω)
The angular velocity \( \omega \) can be calculated from the frequency \( f \):
\[ \omega = 2\pi f \]
Given that the frequency \( f = 50 \, \text{revolutions/second} \):
\[ \omega = 2\pi \times 50 = 100\pi \, \text{radians/second} \]

### Step 4: Calculate the Total Centripetal Force (F)
The formula for centripetal force \( F \) is given by:
\[ F = m r \omega^2 \]
Where:
- \( m \) is the total mass of the chain (2 kg),
- \( r \) is the radius we calculated,
- \( \omega \) is the angular velocity.

Substituting the values:
\[ F = 2 \left(\frac{1.25}{2\pi}\right) (100\pi)^2 \]

### Step 5: Simplify the Expression
Now we simplify the expression:
\[ F = 2 \left(\frac{1.25}{2\pi}\right) (10000\pi^2) \]
\[ F = 2 \times 1.25 \times \frac{10000\pi^2}{2\pi} \]
\[ F = 1.25 \times 10000\pi \]
\[ F = 12500\pi \]

### Step 6: Calculate the Force per Link
Now, we need to find the force on each link. Since there are 125 links:
\[ F_{\text{per link}} = \frac{F}{N} = \frac{12500\pi}{125} \]
\[ F_{\text{per link}} = 100\pi \]

### Step 7: Final Calculation
Using \( \pi \approx 3.14 \):
\[ F_{\text{per link}} = 100 \times 3.14 = 314 \, \text{N} \]

### Conclusion
The centripetal force on each link is approximately **314 N**.

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