Question

A 20 kg child is on a swing that hangs from 3.0-m-long chains. what is her maximum speed if she swings out to a 45° angle?

Answer 1

Answer:

The maximum speed is 4.15 m/s.

Explanation:

To solve this problem we have to use the conservation of energy theorem, which says that the total energy in a system is constant.

In this case, we have kinetic energy (K) and potential energy (U), and the conservation refers to a continuous change between these two while the child is moving. When the child reaches his maximum amplitude on the swing, the kinetic energy is null and the potential energy is maximum. Then, when the child is in the middle of the movement, there the kinetic energy is maximum and the potential energy is null, because there's no height.

It's important to remember that kinetic energy depends on velocity and potential energy depends on position, height, we show this as equations:

$K=\frac{1}{2}mv^{2}$; where $m$ is the mass and $v$ is the velocity.

$U=mgy$; where $m$ is the mass, $g$ is the gravity and $y$ is height.

Now, we analyse the movement. We are gonna use have of the whole movement of the swing, so, the initial movement is gonna be the middle, where the chain is perpendicular to the floor, and the final movement is gonna be where the chain forms 45°. The analysis is transformed into a equation.

So, according to the conservation of energy, the total energy at the beginning of the movement is the same at the final.

$E_{i}=E_{f} \\\frac{1}{2}mv_{i} ^{2}+mgy_{i} =frac{1}{2}mv_{f} ^{2}+mgy_{f}\\\frac{1}{2}mv_{i} ^{2}+0=0+mgy_{f}$

Now we solve for the initial velocity which is also the maximum speed, because we took the initial point where the chain is perpendicular to the ground, and as we said, in that point the potential energy is null and the kinetic energy is max.

$v_{i}=\sqrt{2gy_{f}}$

Now, we need to calculate $y_{f}$:

$cos45=\frac{3-y_{f} }{3} \\3\frac{\sqrt{2}}{2}=3-y_{f} \\y_{f}=3- 3\frac{\sqrt{2}}{2}=0.88$

So, $y_{f}=0.88$, now we replace this to find the initial or maximum velocity:

$v_{i}=\sqrt{2gy_{f}}$

$v_{i}=\sqrt{2(9.8)(0.88)}$

$v_{i}=4.15m/s$

Therefore, the maximum speed is 4.15 m/s.

Answer 2

The solution for this problem is:

Remember that this doesn’t depend on the mass of the child.

E = T + U = constant 
E (maximum height) = T + U =U = mgh = mg[r - r· cos (Θ)] 

E (bottom height) = T + U = T = ½mv² = mg[r - r · cos (Θ)] 

v² = 2g[r – r · cos (Θ)] 

v = √ (2g[r-r·cos(Θ)])

= √(2(9.8)[3 – 3 · cos (45°)])

= 4.15 m/s or 15 kph