A rod standing vertically on a table falls down. What will be the linear velocity of the middle point of the rod at the end of the fall, if the rod is 15 cm long? Assume sufficient friction for no slipping?

Options

A. 1.05m/sec
B. 2.1m/sec
C. 3m/sec
D. 0.1m/sec

Explanation

To find the linear velocity of the middle point of a rod that falls from a vertical position to a horizontal position, we can follow these steps:

### Step 1: Understand the Problem
The rod is 15 cm long and falls from a vertical position. We need to find the linear velocity of the middle point of the rod just before it hits the ground.

### Step 2: Identify Key Points
The center of mass of the rod is located at its midpoint. When the rod falls, it rotates about the point of contact with the table. The height of the center of mass when the rod is vertical is \( \frac{L}{2} \), where \( L \) is the length of the rod.

### Step 3: Calculate the Height of the Center of Mass
Given that the length of the rod \( L = 15 \) cm, the height of the center of mass from the ground when the rod is vertical is:
\[ h = \frac{L}{2} = \frac{15 \text{ cm}}{2} = 7.5 \text{ cm} = 0.075 \text{ m} \]

### Step 4: Apply the Work-Energy Theorem
The work done by gravity when the center of mass falls a height \( h \) is equal to the change in kinetic energy:
\[ W = \Delta KE \]
The work done by gravity is:
\[ W = mgh \]
The change in kinetic energy for the falling rod (initially at rest) is:
\[ \Delta KE = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia of the rod about the point of contact and \( \omega \) is the angular velocity.

### Step 5: Find the Moment of Inertia
The moment of inertia \( I \) of a rod about an axis through one end is given by:
\[ I = \frac{1}{3} m L^2 \]

### Step 6: Relate Angular Velocity to Linear Velocity
The linear velocity \( v \) of the center of mass is related to the angular velocity \( \omega \) by:
\[ v = \omega \frac{L}{2} \]

### Step 7: Combine the Equations
Using the work-energy theorem:
\[ mgh = \frac{1}{2} I \omega^2 \]
Substituting \( I \):
\[ mgh = \frac{1}{2} \left(\frac{1}{3} m L^2\right) \omega^2 \]
Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{1}{6} L^2 \omega^2 \]
Solving for \( \omega \):
\[ \omega^2 = \frac{6gh}{L^2} \]

### Step 8: Substitute for \( v \)
Now substituting \( \omega \) into the equation for \( v \):
\[ v = \omega \frac{L}{2} = \frac{L}{2} \sqrt{\frac{6gh}{L^2}} = \frac{L}{2} \cdot \frac{\sqrt{6gh}}{L} = \frac{\sqrt{6gh}}{2} \]

### Step 9: Substitute Values
Substituting \( g = 9.81 \, \text{m/s}^2 \) and \( h = 0.075 \, \text{m} \):
\[ v = \frac{\sqrt{6 \cdot 9.81 \cdot 0.075}}{2} \]

Calculating this:
\[ v = \frac{\sqrt{4.413}}{2} \approx \frac{2.1}{2} \approx 1.05 \, \text{m/s} \]

### Final Answer
The linear velocity of the middle point of the rod at the end of the fall is approximately \( 1.05 \, \text{m/s} \).

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