To determine the angular velocity of the L-shaped rod at the moment it has rotated through 90 degrees, we can apply the principles of energy conservation. When the rod is released from rest, its potential energy converts into rotational kinetic energy as it falls. Let’s break down the solution step by step.
Understanding the System
We have an L-shaped rod of length 2l, which consists of two segments: one vertical segment and one horizontal segment. The pivot point P is located where these two segments meet. As the rod rotates, we need to find the angular velocity when it reaches the horizontal position.
Energy Considerations
Initially, when the rod is at rest in the vertical position, it has potential energy due to its height. As it rotates, this potential energy is converted into kinetic energy. The total mechanical energy should remain constant, assuming no energy losses due to friction or air resistance.
Calculating Potential Energy
When the rod is vertical, the center of mass of the L-shaped rod is located at a height of l/2 (the center of mass of each segment being at l/2). As it rotates to 90 degrees, the center of mass will drop to a height of 0 (the pivot point). The change in height (h) is:
- Initial height: h_i = l/2
- Final height: h_f = 0
- Change in height: Δh = h_i - h_f = l/2
The potential energy (PE) at the start is given by:
PE_initial = m * g * h_i = m * g * (l/2)
Calculating Kinetic Energy
The kinetic energy (KE) when the rod has rotated through 90 degrees is given by:
KE = (1/2) * I * ω²
where I is the moment of inertia of the rod about the pivot point P and ω is the angular velocity we want to find.
Finding the Moment of Inertia
The moment of inertia I for the L-shaped rod about point P can be calculated as the sum of the moments of inertia of its segments:
- For the vertical segment (length l): I_vertical = (1/3) * m_vertical * l²
- For the horizontal segment (length l): I_horizontal = m_horizontal * (l²) = m_horizontal * (l²)
Since both segments have the same mass m (assuming uniform density), we can express the total moment of inertia as:
I = (1/3)(m/2)(l²) + (m/2)(l²) = (1/3)(m/2)(l²) + (1/2)(m)(l²) = (m * l²) / 2
Applying Conservation of Energy
Now, we can set the initial potential energy equal to the kinetic energy at 90 degrees:
m * g * (l/2) = (1/2) * I * ω²
Substituting I into the equation gives:
m * g * (l/2) = (1/2) * (m * l² / 2) * ω²
Simplifying the Equation
Canceling m from both sides and simplifying, we have:
g * (l/2) = (1/4) * l² * ω²
Multiplying both sides by 4 and rearranging gives:
2g = l * ω²
Therefore, solving for ω yields:
ω² = (2g) / l
Taking the square root:
ω = sqrt(2g/l)
Final Result
In conclusion, the angular velocity of the L-shaped rod at the instant it has rotated through 90 degrees is:
ω = sqrt(2g/l)
This result demonstrates how the gravitational potential energy transforms into kinetic energy during the rod's motion, illustrating the principle of conservation of energy in a rotational system.
