Problem 81 - University Physics with Modern Physics

Question

Rolling Stones. A solid, uniform, spherical boulder starts from rest and rolls down a \(50.0-\mathrm{m}\) -high hill, as shown in Fig. Plo.81. The top half of the hill is rough enough to cause the boulder to roll without slipping, but the lower half is covered with ice and there is no friction. What is the translational speed of the boulder when it reaches the bottom of the hill?

Step-by-Step Solution

Answer

The boulder's translational speed at the bottom is approximately 26.46 m/s.

Step 1: Understand Energy Conservation

The boulder starts with potential energy at the top of the hill and ends with kinetic energy at the bottom. The potential energy at the top is converted to translational and rotational kinetic energy by the time it reaches the bottom of the hill.

Step 2: Initial Potential Energy Calculation

Calculate the gravitational potential energy at the top of the hill using the formula \[PE = mgh\]where \(m\) is the mass of the boulder, \(g = 9.8 \, \text{m/s}^2\) is the acceleration due to gravity, and \(h = 50.0 \, \text{m}\) is the height.

Step 3: Kinetic Energy at the Bottom

At the bottom, the total mechanical energy consists of translational kinetic energy and rotational kinetic energy. Express the kinetic energies as:\[ KE_{trans} = \frac{1}{2} mv^2 \]\[ KE_{rot} = \frac{1}{2} I \omega^2 \]where \( v \) is the translational velocity, and \( \omega \) is the angular velocity. For a sphere, \( I = \frac{2}{5}mr^2 \) and \( \omega = \frac{v}{r} \).

Step 4: Equate Energy Initial and Final States

Using conservation of energy, set up the equation:\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]Substitute \( I = \frac{2}{5}mr^2 \) and \( \omega = \frac{v}{r} \) into the equation:\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} \left( \frac{2}{5} mr^2 \right) \left( \frac{v}{r} \right)^2 \]

Step 5: Simplify and Solve for Translational Speed

Simplify the equation:\[ mgh = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 \]\[ mgh = \frac{7}{10} mv^2 \]Cancel out the mass \(m\) from both sides, and solve for \( v \):\[ gh = \frac{7}{10} v^2 \]\[ v^2 = \frac{10}{7} gh \]\[ v = \sqrt{\frac{10}{7} \times 9.8 \, \text{m/s}^2 \times 50.0 \, \text{m}} \]

Step 6: Calculate and Conclude

Evaluate the expression:\[ v = \sqrt{\frac{10}{7} \times 490} \]\[ v \approx \sqrt{700} \]\[ v \approx 26.46 \, \mathrm{m/s} \]Thus, the translational speed of the boulder at the bottom is approximately 26.46 m/s.

Key Concepts

Potential Energy

Potential energy is a fundamental concept in physics, particularly when studying energy conservation. It is the energy stored in an object due to its position relative to a reference point. For the boulder in our example, this reference point is the top of the hill. The boulder's potential energy at the top is calculated using the formula:

• \( PE = mgh \)

where:

• \(m\) is the mass of the boulder.
• \(g\) is the acceleration due to gravity (approximately 9.8 m/s²).
• \(h\) is the height of the hill (50.0 m, in this case).

Potential energy is what gives the boulder its ability to move or fall down the hill.

As the boulder progresses downward, this stored energy transforms into kinetic energy, both translational and rotational. This transformation aligns with the law of conservation of energy, which states that energy cannot be created or destroyed, only converted from one form to another.

Kinetic Energy

Kinetic energy is the energy of motion. When the boulder rolls down the hill, the potential energy at the top is gradually converted into kinetic energy. At the bottom of the hill, the total kinetic energy consists of:

• Translational kinetic energy, given by the formula \( KE_{trans} = \frac{1}{2} mv^2 \), where \(v\) is the translational velocity.
• Rotational kinetic energy, which arises because the boulder is rolling, given by \( KE_{rot} = \frac{1}{2} I \omega^2 \).

For objects like our boulder, it's essential to consider both forms.

During the boulder's descent, the rough top half of the hill ensures it rolls without slipping, converting energy to both forms of kinetic energy without loss.

At the ice-covered bottom half, the boulder slides without the influence of friction, conserving the already converted kinetic energy to calculate final velocity.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion. It plays a crucial role in determining the boulder's rotational kinetic energy as it rolls down a hill. For our boulder, which is a solid sphere, the moment of inertia is calculated using:

• \( I = \frac{2}{5}mr^2 \)

Here,

• \(m\) is the mass of the boulder.
• \(r\) is its radius.

The moment of inertia affects how the boulder accelerates when rolling.

As the boulder reaches the hill's base, its angular velocity \(\omega\) is connected to the translational velocity \(v\) by the relationship \(\omega = \frac{v}{r}\).

Combined with the moment of inertia, these equations ensure we accurately compute both translational and rotational dynamics in assessing the energy conservation in this exercise.