Problem 57 - College Physics

Question

A hoop starts from rest at a height \(1.2 \mathrm{~m}\) above the base of an inclined plane and rolls down under the influence of gravity. What is the linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface? (Neglect friction.)

Step-by-Step Solution

Answer

The linear speed of the hoop's center of mass as it leaves the incline is approximately 3.43 m/s.

Step 1: Understand the Problem

We need to find the linear speed of the hoop's center of mass when it leaves a frictionless inclined plane and enters a horizontal surface. The hoop starts from rest at a certain height.

Step 2: Identify the Energy Principles

Since the hoop rolls without slipping and friction is neglected, mechanical energy is conserved. This means that the gravitational potential energy (GPE) at the start is converted to kinetic energy (KE) at the bottom.

Step 3: Write the Energy Conservation Equation

At the top, the hoop has gravitational potential energy \( \text{GPE} = mgh \), where \( m \) is mass, \( g = 9.81 \text{ m/s}^2 \), and \( h = 1.2 \text{ m} \). At the bottom, the energy is converted to translational and rotational kinetic energy: \( KE = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 \), where \( v \) is linear speed and \( I = mr^2 \) (moment of inertia for hoop) and \( \omega = \frac{v}{r} \).

Step 4: Simplify the Equation

Substituting \( I = mr^2 \) and \( \omega = \frac{v}{r} \) into the kinetic energy yields \( KE = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 = mv^2 \). Thus, the total kinetic energy is \( mv^2 \).

Step 5: Solve for Linear Speed

Setting the initial gravitational potential energy equal to the kinetic energy at the bottom, \( mgh = mv^2 \). Cancel out mass \( m \), resulting in \( gh = v^2 \). This gives us \( v = \sqrt{gh} = \sqrt{9.81 \text{ m/s}^2 \times 1.2 \text{ m}} \).

Step 6: Calculate the Linear Speed

Calculate the numeric value of \( v \): \( v = \sqrt{11.772} \approx 3.43 \text{ m/s} \).

Key Concepts

Gravitational Potential Energy

Gravitational Potential Energy (GPE) is the energy stored in an object due to its position in a gravitational field. For an object elevated at a height, this energy depends primarily on three factors: mass, gravitational acceleration, and height above the ground.
When calculating GPE, the formula used is given by:

• \( ext{GPE} = mgh \),

where \( m \) is the mass of the object, \( g \) stands for the acceleration due to gravity (approximately \( 9.81\, \text{m/s}^2 \) on Earth), and \( h \) is the height.
In our specific example of a hoop rolling down an inclined plane, the GPE at the top of the incline is transferred into kinetic energy as it rolls down. This transformation of energy is essential to understanding how the speed of the hoop is determined at the bottom of the incline.

Kinetic Energy

Kinetic Energy (KE) refers to the energy a body possesses due to its motion. It comes in two forms: translational and rotational.
Translational kinetic energy relates to the linear motion of an object's center of mass and is calculated using:

• \( ext{KE}_{ ext{trans}} = \frac{1}{2} mv^2 \),

where \( m \) is the mass of the object and \( v \) is its linear velocity.
Rotational kinetic energy, on the other hand, refers to the energy from spinning and is given by:

• \( ext{KE}_{ ext{rot}} = \frac{1}{2} I \omega^2 \),

where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
For the hoop on the inclined plane, both forms of kinetic energy are considered. The total kinetic energy when the hoop rolls off the incline is the sum of these two types.

Inclined Plane

An inclined plane is a simple machine consisting of a flat surface tilted at an angle to the horizontal. It allows objects to be moved up or down with less force than lifting them vertically.
When analyzing motion on an inclined plane, gravitational force plays a significant role, contributing to the acceleration of an object down the slope.
Key factors that influence the motion on an inclined plane include:

• The angle of the incline, which affects the component of gravitational force driving the object down the plane.
• The nature of the surface and presence of friction, although, in our hoop example, friction is ignored.

In the provided exercise, the hoop descends a frictionless inclined plane, converting its gravitational potential energy entirely into kinetic energy as it reaches the bottom.

Rotational Motion

Rotational Motion is the movement of an object around a center or an axis. This concept is particularly relevant to objects like hoops or wheels that roll.
The two primary components of rotational motion are:

• The moment of inertia \( I \), which is the rotational equivalent of mass and depends on the distribution of mass around the axis of rotation.
• Angular velocity \( \omega \), the rate at which an object rotates around an axis, typically measured in radians per second.

For the hoop in the exercise, the formula for relating linear velocity \( v \) to angular velocity is \( \omega = \frac{v}{r} \), where \( r \) is the radius of the hoop.
The hoop's rotational motion comes into play as gravitational energy transforms into both translational and rotational kinetic energy by the time it exits the incline.