Problem 83 - University Physics with Modern Physics

Question

You are designing a slide for a water park. In a sitting position, park guests slide a vertical distance \(h\) down the waterslide, which has negligible friction. When they reach the bottom of the slide, they grab a handle at the bottom end of a 6.00-m-long uniform pole. The pole hangs vertically, initially at rest. The upper end of the pole is pivoted about a stationary, frictionless axle. The pole with a person hanging on the end swings up through an angle of 72.0\(^\circ\), and then the person lets go of the pole and drops into a pool of water. Treat the person as a point mass. The pole's moment of inertia is given by \(I = \frac{1}{3} ML^2\), where \(L =\) 6.00 m is the length of the pole and \(M =\) 24.0 kg is its mass. For a person of mass 70.0 kg, what must be the height h in order for the pole to have a maximum angle of swing of 72.0\(^\circ\) after the collision?

Step-by-Step Solution

Answer

The required height h is approximately 3.64 meters.

Step 1: Understand the Conservation of Mechanical Energy

When the person slides down the waterslide, the potential energy lost is equal to the kinetic energy gained at the bottom of the slide. This conversion can be expressed as: \[ mgh = \frac{1}{2} mv^2 \]where \(m\) is the mass of the person, \(g\) is acceleration due to gravity (9.81 m/s²), \(h\) is the height, and \(v\) is the speed of the person at the bottom of the slide.

Step 2: Analyze the Collision with the Pole

When the person grabs the pole, conservation of angular momentum applies because no external torques are acting. The angular momentum of the person about the pivot point just before the collision is \(m \cdot v \cdot L\), where \(v\) is the speed of the person and \(L\) is the length of the pole.This is equal to the angular momentum of the system just after the collision:\[ m \cdot v \cdot L = I \cdot \omega + m \cdot L^2 \cdot \omega \]Substitute \(I = \frac{1}{3}ML^2\):\[ m \cdot v \cdot L = \left( \frac{1}{3}ML^2 + mL^2 \right) \omega \]

Step 3: Analyze Maximum Angle Using Conservation of Energy

After the person starts swinging with the pole, their potential energy at the maximum angle is: \[ (M + m)gh' \]where \(h' = L(1 - \cos\theta)\). The initial kinetic energy just before swinging is converted into potential energy:\[ \frac{1}{2} I \omega^2 + \frac{1}{2} mL^2\omega^2 = (M + m)gh' \]Using small angle approximations or detailed calculation, find \(\omega\) and solve to find \(h\).

Step 4: Calculate Required Height h

Find \(\omega\) by combining equations from Step 2 and 3. Solve the equations simultaneously to find the speed \(v\) and then calculate \(h\) using:\[ h = \frac{v^2}{2g} \]Given that the pole swings to 72 degrees, calculate \(v\) considering that energy at the peak is kinetic energy converted into potential.

Key Concepts

Conservation of Angular Momentum

When a person reaches the bottom of the slide and grabs the pole, they introduce a critical concept from physics - the conservation of angular momentum. This principle is pivotal because no external torques act on the system, which means that the total angular momentum before the interaction is equal to the total angular momentum afterward.
The angular momentum of a moving object is proportional to its mass, its velocity, and the distance from the pivot point. This can be expressed as \( m \cdot v \cdot L \), where \(m\) is the mass, \(v\) is the velocity, and \(L\) is the length of the pole.
After the collision, this angular momentum is transferred into the rotational motion of the pole along with the person now holding onto it. The equation reflecting this balance is:

• \( m \cdot v \cdot L = \left( \frac{1}{3}ML^2 + mL^2 \right) \omega \)

Here, \( \omega \) is the angular velocity that needs to be solved for. The key takeaway is that the angular momentum "spreads out" into the entire pole and person system, illustrating the conservation law.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotation. For the pole, which rotates about one end, it’s crucial to understand how this inertia influences the dynamics when the person grabs onto it.
For a rigid body such as the uniform pole, the moment of inertia \( I \) is calculated by the formula \( I = \frac{1}{3}ML^2 \), where \(M\) is the mass of the pole and \(L\) is its length. This formula takes into account the distribution of mass along the length of the pole.

• The higher the moment of inertia, the harder it is to spin the object.
• Adding another mass at a certain distance (like the person at the end) also contributes to the total inertia.

Thus, when applying forces or when rotational motion is involved as is the case with swinging, the moment of inertia dictates how quickly or easily the object can change its state of motion.

Kinematic Equations

The problem of the water park slide can be further understood using kinematic equations, especially when analyzing the motion of the person and the pole. Kinematics involves the description of motion using parameters like velocity, acceleration, and displacement.
While the person slides down the water slide, they accelerate due to gravity, transforming potential energy into kinetic energy, as described by \( mgh = \frac{1}{2} mv^2 \). This equation explains that the energy at the top of the slide is effectively transferred into speed at the bottom.

• \( h \) represents height, connected to potential energy.
• \( v \) is the velocity at the bottom, vital for determining initial conditions for rotation.
• Kinematic equations also help calculate the maximum height achieved along the circular path after grabbing the pole.

These equations demonstrate how foundational physics principles, like energy conversion, impact the motion observed in this practical exercise.