Problem 48 - College Physics

Question

A solid uniform marble and a block of ice, each with the same mass, start from rest at the same height \(H\) above the bottom of a hill and move down it. The marble rolls without slipping, but the ice slides without friction. (a) Find the speed of each of these objects when it reaches the bottom of the hill. (b) Which object is moving faster at the bottom, the ice or the marble? (c) Which object has more kinetic energy at the bottom, the ice or the marble?

Step-by-Step Solution

Answer

(a) Ice: \( \sqrt{2gh} \), Marble: \( \sqrt{\frac{10}{7}gh} \); (b) Ice; (c) Equal.

Step 1: Understanding the Problem

We have a marble that rolls down the hill and a block of ice that slides down the hill. We need to determine their speeds and kinetic energies at the bottom of the hill. The marble has rotational kinetic energy due to rolling, while the block of ice, which slides without friction, only has translational kinetic energy.

Step 2: Calculate Translational Speed of the Block of Ice

Since the block of ice slides down the hill without friction, it converts all of its gravitational potential energy into translational kinetic energy at the bottom. Using the conservation of energy:\[ mgh = \frac{1}{2} mv^2 \]Solving for \( v \):\[ v = \sqrt{2gh} \]

Step 3: Calculate Translational Speed of the Marble

For the marble, rolling without slipping means it has both translational and rotational kinetic energy. Using conservation of energy:\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 \]For a solid sphere, \( I = \frac{2}{5}mr^2 \) and \( v = r\omega \). Substituting these into the equation:\[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5}mr^2\right) \left(\frac{v^2}{r^2}\right) \]Simplifying gives:\[ mgh = \frac{1}{2} mv^2 + \frac{1}{5} mv^2 \]\[ mgh = \frac{7}{10} mv^2 \]Solving for \( v \):\[ v = \sqrt{\frac{10}{7}gh} \]

Step 4: Compare the Speeds of the Marble and Ice

The speed of the block of ice is \( \sqrt{2gh} \), and the speed of the marble is \( \sqrt{\frac{10}{7}gh} \). Comparing:Since \( \frac{10}{7} \approx 1.4286 \) and \( 2 = 2.0 \), we see that \( \sqrt{2gh} > \sqrt{\frac{10}{7}gh} \). Therefore, the block of ice is moving faster at the bottom.

Step 5: Determine Which Object Has More Kinetic Energy

The gravitational potential energy lost is the same for both objects, which means their total mechanical energy at the bottom is also the same. While the ice convert this completely into translational kinetic energy, the marble distributes it between translational and rotational energies. Thus, since they start with the same potential energy, they have the same *total* kinetic energy at the bottom.

Key Concepts

Translational Speed

Translational speed refers to the speed at which the center of mass of an object moves along a path. In the scenario with the marble and the block of ice, their translational speeds at the bottom are determined by how their gravitational potential energy is converted into kinetic energy.
For the block of ice, as it slides down the hill without friction, its gravitational potential energy is fully transformed into translational kinetic energy. The formula for translational speed is derived from the conservation of energy principle:

• Potential Energy: \(mgh\)
• Translational Kinetic Energy: \(\frac{1}{2} mv^2\)

Setting these equal gives us \(mgh = \frac{1}{2} mv^2\), and solving for \(v\), the translational speed of the ice block becomes \(v = \sqrt{2gh}\). Every bit of energy turns into translational motion, making it straightforward to determine how fast something like the ice block moves at the hill's bottom.
For the marble, both translational and rotational kinetic energy need to be taken into account, which affects how we calculate its translational speed.

Rotational Kinetic Energy

Rotational kinetic energy is the energy an object possesses due to its rotation. For any rolling object, such as our marble in this problem, it contributes significantly alongside translational kinetic energy. Here, the marble rolls without slipping, meaning it directly converts part of its initial gravitational potential energy into rotational kinetic energy.
The formula that represents this form of energy for a rotating sphere like a marble is:

• Rotational Kinetic Energy: \(\frac{1}{2} I \omega^2\)

Where \(I\) is the moment of inertia for a solid sphere \(\left(\frac{2}{5}mr^2\right)\), and \(\omega\) is the angular velocity. In rolling scenarios, there is a relationship between the translational velocity \(v\) and angular velocity \(\omega\), which is \(v = r\omega\). Substituting the values and simplifying:

• Moment of Inertia: \(\frac{2}{5}mr^2\)
• \(mgh = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{2}{5}mr^2\right)\left(\frac{v^2}{r^2}\right)\)

This leads to calculating the marble's speed as \(v = \sqrt{\frac{10}{7}gh}\). The presence of rotational kinetic energy reduces the translational speed slightly compared to the block of ice.

Gravitational Potential Energy

Gravitational potential energy is the energy an object holds due to its position in a gravitational field. It's particularly vital in scenarios like the rolling marble and sliding ice block as it represents their starting energy at height \(H\). This potential energy is given by:

• Gravitational Potential Energy: \(mgh\)

As both the marble and ice slide from rest at height \(H\), their potential energy at the top is converted entirely to kinetic energy at the bottom. What makes this interesting is how each object utilizes this energy.

• The ice block converts all of its potential energy into translational kinetic energy, resulting in \(v = \sqrt{2gh}\).
• The marble splits its energy into translational and rotational forms, thus the translational speed \(v = \sqrt{\frac{10}{7}gh}\) reflects its dual energy usage.

Even though both objects start with the same gravitational potential energy, they experience different speeds and energy types at the bottom due to the rolling motion of the marble.