Questions: Part A A ring, a disk, and a solid sphere begin rolling down a hill together. Which reaches the bottom first? all reach the bottom at the same time sphere ring disk need more information Submit Request Answer

Solution Steps

The problem involves three objects: a ring, a disk, and a solid sphere, all rolling down a hill. We need to determine which object reaches the bottom first. This is a classic physics problem involving rotational motion and energy conservation.

When objects roll without slipping, their motion is a combination of translational and rotational motion. The key factor that determines which object reaches the bottom first is the moment of inertia, which affects how the gravitational potential energy is converted into kinetic energy.

The moment of inertia \(I\) for each object is different:

• For a ring: \(I = mR^2\)
• For a disk: \(I = \frac{1}{2}mR^2\)
• For a solid sphere: \(I = \frac{2}{5}mR^2\)

Where \(m\) is the mass and \(R\) is the radius of the object.

The total mechanical energy for each object is conserved. The potential energy at the top is converted into translational and rotational kinetic energy at the bottom: \[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \]

Since \(\omega = \frac{v}{R}\), we can substitute and solve for \(v\), the linear velocity at the bottom: \[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\left(\frac{v}{R}\right)^2 \]

The object with the smallest moment of inertia relative to its mass will have the highest velocity at the bottom, as more of its potential energy is converted into translational kinetic energy.

• For the ring: \(v = \sqrt{\frac{2gh}{1 + 1}} = \sqrt{gh}\)
• For the disk: \(v = \sqrt{\frac{2gh}{1 + \frac{1}{2}}} = \sqrt{\frac{4gh}{3}}\)
• For the solid sphere: \(v = \sqrt{\frac{2gh}{1 + \frac{2}{5}}} = \sqrt{\frac{10gh}{7}}\)

The object with the highest velocity reaches the bottom first. Comparing the velocities:

• The solid sphere has the highest velocity, followed by the disk, and then the ring.

Final Answer