To solve the problem of finding the linear velocity of the middle point of a vertically standing rod that falls down, we can follow these steps:
### Step 1: Understand the Initial Setup
The rod is initially standing vertically on a table and has a length of 15 cm. When it falls, the bottom end of the rod remains in contact with the table due to sufficient friction, which means that the rod will rotate about this point as it falls.
### Step 2: Identify the Center of Mass
The center of mass of the rod is located at its midpoint. For a rod of length \( L = 15 \, \text{cm} \), the height of the center of mass from the ground when the rod is vertical is:
\[
h_{\text{initial}} = \frac{L}{2} = \frac{15 \, \text{cm}}{2} = 7.5 \, \text{cm} = 0.075 \, \text{m}
\]
### Step 3: Calculate Initial Potential Energy
The initial potential energy (U_initial) of the rod when it is vertical can be calculated using the formula:
\[
U_{\text{initial}} = mgh_{\text{initial}} = mg \cdot 0.075
\]
where \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).
### Step 4: Calculate Final Potential Energy
When the rod falls and is horizontal, the height of the center of mass from the ground is zero. Therefore, the final potential energy (U_final) is:
\[
U_{\text{final}} = 0
\]
### Step 5: Calculate Initial Kinetic Energy
Initially, the rod is at rest, so the initial kinetic energy (K_initial) is:
\[
K_{\text{initial}} = 0
\]
### Step 6: Calculate Final Kinetic Energy
When the rod is horizontal, it is rotating about the bottom end. The final kinetic energy (K_final) can be expressed as:
\[
K_{\text{final}} = \frac{1}{2} I \omega^2
\]
where \( I \) is the moment of inertia of the rod about the pivot point (bottom end) and \( \omega \) is the angular velocity.
For a rod of length \( L \) rotating about one end:
\[
I = \frac{1}{3} mL^2
\]
### Step 7: Apply Conservation of Energy
Using the principle of conservation of energy, we have:
\[
U_{\text{initial}} + K_{\text{initial}} = U_{\text{final}} + K_{\text{final}}
\]
Substituting the values:
\[
mg \cdot 0.075 + 0 = 0 + \frac{1}{2} \left(\frac{1}{3} mL^2\right) \omega^2
\]
This simplifies to:
\[
mg \cdot 0.075 = \frac{1}{6} mL^2 \omega^2
\]
### Step 8: Solve for Angular Velocity
Cancelling \( m \) from both sides and rearranging gives:
\[
g \cdot 0.075 = \frac{1}{6} L^2 \omega^2
\]
\[
\omega^2 = \frac{6g \cdot 0.075}{L^2}
\]
Substituting \( L = 0.15 \, \text{m} \):
\[
\omega^2 = \frac{6 \cdot 9.8 \cdot 0.075}{(0.15)^2}
\]
Calculating this gives:
\[
\omega^2 = \frac{6 \cdot 9.8 \cdot 0.075}{0.0225} = \frac{4.41}{0.0225} \approx 196
\]
\[
\omega \approx \sqrt{196} \approx 14 \, \text{rad/s}
\]
### Step 9: Calculate Linear Velocity of the Midpoint
The linear velocity \( V \) of the midpoint of the rod can be calculated using:
\[
V = \frac{L}{2} \cdot \omega
\]
Substituting the values:
\[
V = \frac{0.15}{2} \cdot 14 = 0.075 \cdot 14 \approx 1.05 \, \text{m/s}
\]
### 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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