Problem 50 - 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

The ice moves faster at the bottom. Both have equal kinetic energy at the bottom.

Step 1: Understand the Problem

We have two objects: a marble that rolls down without slipping and a block of ice that slides down without friction. Both start with the same mass and height, and we need to find their speeds at the bottom and compare their kinetic energies.

Step 2: Apply the Conservation of Energy for the Ice Block

Since the ice block slides without friction, only translational kinetic energy is considered. Using conservation of energy:\[ mgh = \frac{1}{2} mv^2 \]where \( m \) is mass, \( g \) is gravitational acceleration, \( h \) is the height, and \( v \) is velocity. Solve for \( v \):\[v = \sqrt{2gh}\]

Step 3: Apply the Conservation of Energy for the Rolling Marble

The marble rolls, so it has both translational and rotational kinetic energy. The conservation of energy equation becomes: \[mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2\]where \( I = \frac{2}{5}mr^2 \) for a solid sphere and \( \omega = \frac{v}{r} \). Substitute and solve for \( v \): \[mgh = \frac{1}{2}mv^2 + \frac{1}{2} \left(\frac{2}{5}mr^2\right) \left(\frac{v}{r}\right)^2 \]\[mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 \]\[mgh = \frac{7}{10}mv^2 \]\[v = \sqrt{\frac{10gh}{7}}\]

Step 4: Compare the Speeds

Compare the speeds derived: - Ice block speed: \( \sqrt{2gh} \)- Marble speed: \( \sqrt{\frac{10gh}{7}} \)Since \( \sqrt{2gh} > \sqrt{\frac{10gh}{7}} \), the ice is moving faster at the bottom.

Step 5: Compare the Kinetic Energies

Kinetic energy of the ice block:\[ KE_{ice} = \frac{1}{2}mv^2 = \frac{1}{2}m(\sqrt{2gh})^2 = mgh \]Kinetic energy of the marble:\[ KE_{marble} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh \]Both have the same kinetic energy, \( mgh \), because all initial potential energy is converted to kinetic energy. Hence, they have equal kinetic energy at the bottom.

Key Concepts

Translational Kinetic Energy

Translational kinetic energy is a form of energy related to the motion of an object moving along a path. It exclusively accounts for motion in a straight line.
In the case of a block of ice sliding down a hill without friction, all gravitational potential energy is converted into translational kinetic energy. This is expressed by the conservation of energy equation:

• Gravitational potential energy = translational kinetic energy
• \[ mgh = \frac{1}{2} mv^2 \]

Here, \(m\) is mass, \(g\) is gravitational acceleration, \(h\) is height, and \(v\) is velocity.
At the bottom of the hill, the ice block reaches a velocity of \(v = \sqrt{2gh}\).
It is interesting to note that since there is no friction involved, all potential energy transforms into translational kinetic energy, making calculations more straightforward.

Rotational Kinetic Energy

Rotational kinetic energy arises when an object spins around an axis. For a rolling sphere like a uniform marble, both translational and rotational kinetic energies must be considered.
As the marble rolls down the hill, the energy conserved can be described as:

• Total kinetic energy = translational kinetic energy + rotational kinetic energy
• \[ mgh = \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2\]
• where \( I = \frac{2}{5}mr^2 \) and \( \omega = \frac{v}{r} \) for a solid sphere

The expression simplifies to:

• \[ mgh = \frac{7}{10} mv^2 \]
• yielding a final speed of \(v = \sqrt{\frac{10gh}{7}}\)

This showcases how rolling motion involves both sliding along the ground and rotating around its center, splitting the energy between both forms.

Uniform Marble

A uniform marble is an object with identical material composition and mass distribution throughout. This uniformity is crucial as it affects how the marble rotates and how energy is distributed between translational and rotational kinetic forms.
For a solid sphere such as a marble, the moment of inertia is given by:

• \[ I = \frac{2}{5} mr^2\]

This represents the resistance to change in rotational motion.
The combination of these factors determines how much energy ends up as rotational kinetic energy versus translational kinetic energy, impacting the marble's speed compared to the sliding block of ice. This link between moment of inertia, mass distribution, and energy conversion is key to understanding how different shapes and compositions affect motion.

Frictionless Motion

Frictionless motion implies no opposing force that would convert kinetic energy into other forms of energy.
The ice block sliding down the hill illustrates this concept perfectly, lacking friction to slow it down and only using gravitational force to change its state.
In an idealized scenario, all gravitational potential energy converts to translational kinetic energy:

• Gravitational potential equals kinetic energy at the hill bottom
• \[ mgh = \frac{1}{2} mv^2\]

This setup helps students understand simple energy conversion cases.
It removes complicating factors like friction, making calculations more straightforward and focusing learning on fundamental physics principles. Such frictionless setups often serve educational purposes by simplifying scenarios to their core elements.