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A rod of length L is hinged from one end...

A rod of length L is hinged from one end. It is brought to a horizontal position and released. The angular velocity of the rod, when it is in vertical position, is

A

`sqrt((2g)/L)`

B

`sqrt((3g)/L)`

C

`sqrt(g/(2L))`

D

`sqrt(g/l)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the angular velocity of a rod of length \( L \) when it is in a vertical position after being released from a horizontal position, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the System**: - We have a uniform rod of length \( L \) hinged at one end. When the rod is in the horizontal position, it is released and swings down to the vertical position. 2. **Identify the Energy Changes**: - When the rod is released from the horizontal position, it loses gravitational potential energy as it falls to the vertical position. This lost potential energy is converted into rotational kinetic energy. 3. **Calculate the Loss in Potential Energy**: - The center of mass of the rod is located at a distance of \( \frac{L}{2} \) from the hinge. - The initial height of the center of mass when the rod is horizontal is \( \frac{L}{2} \) (since it is at the same level as the hinge). - When the rod is vertical, the center of mass is at the hinge level (height = 0). - Therefore, the loss in gravitational potential energy (PE) is given by: \[ \Delta PE = mgh = mg \left(\frac{L}{2}\right) \] 4. **Calculate the Gain in Rotational Kinetic Energy**: - The rotational kinetic energy (KE) of the rod when it is vertical is given by: \[ KE = \frac{1}{2} I \omega^2 \] - For a rod rotating about one end, the moment of inertia \( I \) is: \[ I = \frac{1}{3} m L^2 \] 5. **Set Up the Energy Conservation Equation**: - According to the conservation of energy: \[ \Delta PE = KE \] - Substituting the expressions we derived: \[ mg \left(\frac{L}{2}\right) = \frac{1}{2} \left(\frac{1}{3} m L^2\right) \omega^2 \] 6. **Simplify the Equation**: - Cancel \( m \) from both sides (assuming \( m \neq 0 \)): \[ g \left(\frac{L}{2}\right) = \frac{1}{6} L^2 \omega^2 \] - Rearranging gives: \[ \omega^2 = \frac{6g}{L} \] 7. **Take the Square Root**: - To find \( \omega \): \[ \omega = \sqrt{\frac{6g}{L}} \] ### Final Answer: The angular velocity \( \omega \) of the rod when it is in the vertical position is: \[ \omega = \sqrt{\frac{6g}{L}} \]
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