To tackle this problem, we need to analyze the motion of the swing using principles from physics, particularly energy conservation and circular motion. Let’s break it down step by step.
Understanding the Swing's Motion
The swing can be modeled as a pendulum. When the child is pulled back to an angle of one radian, the swing is at a height above its lowest point. When the father pushes the child and then releases him at the vertical position, we can determine how high the swing will go based on the energy involved.
Part A: Determining the Maximum Height
First, we need to calculate the initial height of the swing when it is pulled back to an angle of one radian. The height \( h \) can be found using the formula:
- Height above the lowest point: \( h = l - l \cos(\theta) \)
Here, \( l \) is the length of the rope, and \( \theta \) is the angle with the vertical (1 radian). Thus, we can substitute:
- Using \( \cos(1) \approx 0.5403 \):
- Height \( h = l - l \cdot 0.5403 = l(1 - 0.5403) \approx 0.4597l \)
Next, when the father pushes the child with a force \( F = mg \) along the arc, he does work on the child. The work done by the father is equal to the change in potential energy as the swing rises. The potential energy at the maximum height \( H \) can be expressed as:
- Potential Energy \( PE = mgh \)
At the maximum height, the total energy is conserved. The initial potential energy when pulled back is converted into potential energy at the maximum height:
- Setting the energies equal: \( mg(0.4597l) + 0 = mgH \)
From this, we can solve for \( H \):
Part B: Calculating the Duration of the Push
To find out how long the father pushed the child, we need to consider the motion along the arc of the swing. The force applied by the father is equal to the gravitational force acting on the child, which means the child is pushed with a force equal to their weight.
The acceleration \( a \) of the child can be determined using Newton's second law:
- Force \( F = ma \Rightarrow a = \frac{F}{m} = g \text{ (since } F = mg\text{)}\)
Now, we can find the time it takes to travel along the arc from the angle of one radian back to the vertical position. The arc length \( s \) can be calculated as:
- Arc length \( s = l \theta = l \cdot 1 = l \)
Using the kinematic equation for constant acceleration, we can find the time \( t \) it takes to travel this distance:
- Using \( s = ut + \frac{1}{2}at^2 \) (initial velocity \( u = 0 \)):
- Thus, \( l = 0 + \frac{1}{2}gt^2 \Rightarrow t^2 = \frac{2l}{g} \Rightarrow t = \sqrt{\frac{2l}{g}} \)
Final Results
In summary, the swing will rise to a height of approximately \( 0.4597l \) after being pushed, and the time the father pushed the child is given by \( t = \sqrt{\frac{2l}{g}} \). This analysis combines concepts of energy conservation and kinematics to provide a comprehensive understanding of the swing's motion.
