A child of mass m sits in a swing of negligible mass suspended by a rope of length l. Assume that the dimensions of the child are negligible compared with l.
His father pulls the child back until the rope makes an angle of one radian with vertical, then pushes with a force F=mg along the arc of the circle releasing at the vertical :
a) How high up will the swing go?
b) How long did the father push?

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

When the child is pulled back to an angle of one radian, we can determine the height the swing will reach after being released. The swing behaves like a pendulum, and we can use the conservation of mechanical energy to find the maximum height.

Part A: Maximum Height Calculation

Initially, when the child is at an angle of one radian, the potential energy (PE) can be calculated based on the height gained from the lowest point of the swing. The height (h) can be found using the formula:

• h = l - l * cos(θ)

Here, θ is the angle in radians, and l is the length of the rope. Plugging in θ = 1 radian:

• h = l - l * cos(1)
• h = l(1 - cos(1))

Next, we convert this height into potential energy:

• PE_initial = mgh = mg(l(1 - cos(1)))

When the swing reaches its highest point after being pushed, all this potential energy will convert into kinetic energy (KE) at the lowest point:

• KE = (1/2)mv^2

At the lowest point, the potential energy is zero, and the kinetic energy is at its maximum. Setting the initial potential energy equal to the kinetic energy gives:

• mg(l(1 - cos(1))) = (1/2)mv^2

We can cancel out the mass (m) since it appears on both sides:

• g(l(1 - cos(1))) = (1/2)v^2

Solving for v, we find:

• v = sqrt(2g(l(1 - cos(1))))

Now, when the swing is released from the vertical position, it will rise to a height where all kinetic energy converts back to potential energy. The maximum height (H) reached can be calculated as:

• PE_final = mgh = (1/2)mv^2

Setting these equal gives:

• mgH = (1/2)mv^2

Again, we can cancel m:

• gH = (1/2)v^2

Substituting for v from earlier:

• gH = (1/2)(2g(l(1 - cos(1))))

Simplifying this, we find:

• H = l(1 - cos(1))

Thus, the swing will rise to a height of:

• H = l(1 - cos(1))

Part B: Duration of the Push

To determine how long the father pushed the swing, we need to consider the dynamics of the swing's motion. The force applied by the father (F = mg) acts along the arc of the circle. The swing moves in a circular path, and we can find the time taken to push the swing to the vertical position using the concept of angular motion.

The angular displacement (θ) during the push is one radian. The tangential acceleration (a_t) due to the force F can be calculated as:

• a_t = F/m = g

Using the relationship between tangential acceleration and angular acceleration (α), we have:

• a_t = rα
• g = lα

Thus, the angular acceleration is:

• α = g/l

Using the kinematic equation for angular motion:

• θ = ω_initial * t + (1/2)αt^2

Assuming the initial angular velocity (ω_initial) is zero, we can simplify this to:

• 1 = (1/2)(g/l)t^2

Solving for t gives:

• t^2 = (2l/g)
• t = sqrt(2l/g)

In summary, the time the father pushed the swing is:

• t = sqrt(2l/g)

In conclusion, the swing will rise to a height of l(1 - cos(1)), and the duration of the push will be sqrt(2l/g).