To tackle the problem of finding the acceleration of the center of mass of a thin uniform rod that is free to rotate about a horizontal axis, we need to first understand the system's dynamics. In this scenario, we have a rod of mass \(m\) and length \(4l\) that is pivoting from a point \(l\) away from one end. When the rod is released from a horizontal position, it begins to rotate due to the force of gravity acting on its center of mass. Let's break down the steps to calculate the acceleration of the center of mass at that instant.
Identifying Key Components
1. **Center of Mass Location**: For a uniform rod, the center of mass is located at its midpoint. Given that the total length of the rod is \(4l\), the center of mass is positioned at \(2l\) from one end.
2. **Distance from Pivot**: Since the pivot is located \(l\) from one end, the distance from the pivot to the center of mass is \(2l - l = l\).
Forces Acting on the Rod
When the rod is released, two primary forces act on it:
- The weight of the rod (\(mg\)), acting downward at the center of mass.
- The support reaction at the pivot, which provides the necessary torque for rotation.
Calculating Torque
Next, we calculate the torque (\(\tau\)) about the pivot point. The torque is given by the equation:
\(\tau = r \times F\)
Where \(r\) is the distance from the pivot to the center of mass (which we found to be \(l\)), and \(F\) is the force acting at that point (the weight of the rod).
Substituting the values, we have:
\(\tau = l \times mg\)
Moment of Inertia
To find the angular acceleration (\(\alpha\)) of the rod, we also need the moment of inertia (\(I\)) of the rod about the pivot point. For a thin rod rotating about an axis not through its center, the moment of inertia is given by:
\(I = \frac{1}{3} m L^2\)
In our case, \(L = 4l\), so:
\(I = \frac{1}{3} m (4l)^2 = \frac{16}{3} ml^2\)
Applying Newton's Second Law for Rotation
According to Newton's second law for rotation, the net torque is equal to the moment of inertia times the angular acceleration:
\(\tau = I \alpha\)
Substituting the values we have:
\(l \cdot mg = \frac{16}{3} ml^2 \cdot \alpha\)
Solving for Angular Acceleration
Canceling \(m\) from both sides and rearranging gives:
\(\alpha = \frac{3g}{16l}\)
Acceleration of the Center of Mass
The linear acceleration (\(a_{cm}\)) of the center of mass is related to the angular acceleration by the formula:
\(a_{cm} = r \cdot \alpha\)
Here, \(r\) is the distance from the pivot to the center of mass, which we found to be \(l\). Thus:
\(a_{cm} = l \cdot \frac{3g}{16l} = \frac{3g}{16}\)
Final Result
The acceleration of the center of mass of the rod at the instant it is released is:
\(a_{cm} = \frac{3g}{16}\)
This calculation shows how rotational dynamics and linear motion principles are interconnected. Understanding these relationships is crucial for analyzing similar systems in physics.
