Problem 15 - Classical Mechanics - An Undergraduate Text

Question

A uniform ball of radius \(a\) can roll without slipping on the outside surface of a fixed sphere of (outer) radius \(b\) and centre \(O\). Initially the ball is at rest at the highest point of the sphere when it is slightly disturbed. Find the speed of the centre \(G\) of the ball in terms of the variable \(\theta\), the angle between the line \(O G\) and the upward vertical. [Assume planar motion.]

Step-by-Step Solution

Answer

The speed of the center of the ball in terms of the variable \(\theta\) is given by \( v = \sqrt{ \frac{10g}{7} [b\cos(\theta) - a] } \)

Step 1: Understand the Situation

The ball is initially at rest at the top of the sphere. it will start rolling down due to gravitational pull. As it descends, it gains kinetic energy while losing potential energy.

Step 2: Determine potential energy

The gravitational potential energy of the ball is given by \( mgh \), where \( m \) is the mass of the ball, \( g \) is the acceleration due to gravity, and \( h \) is the height from the bottom of the sphere to the center of the ball. The height can be expressed in terms of \( b \), \( a \), and \( \theta \) as \( h = b\cos(\theta) - a\). That makes the potential energy \( PE = mgb\cos(\theta) - mga \).

Step 3: Find the kinetic energy

The kinetic energy consists of translational and rotational parts; for rolling without slipping, this is \( K = \frac{1}{2}mv^2 + \frac{1}{2}Iw^2 \), where \( v \) is the linear speed, \( I \) the moment of inertia and \( w \) the rotational velocity. For a rolling sphere, \( I = \frac{2}{5}mR^2 \), and \( v = wR \), which gives us \( K = \frac{7}{10}mv^2 \).

Step 4: Apply conservation of mechanical energy

Since there are no non-conservative forces, the total mechanical energy of the ball is conserved; this means that the kinetic energy plus the potential energy is constant: \( PE_{i} + K_{i} = PE_{f} + K_{f} \). On substituting and simplifying, we get the final equation \( v = \sqrt{ \frac{10g}{7} [b\cos(\theta) - a] } \).

Key Concepts

Conservation of Mechanical Energy

Understanding the conservation of mechanical energy is crucial when analyzing the motion of objects. Simply put, this principle tells us that in the absence of non-conservative forces, such as friction or air resistance, the total mechanical energy of a system remains constant. Mechanical energy itself is the sum of kinetic energy, which is energy due to motion, and potential energy, which is stored energy due to an object's position or arrangement.

For a ball rolling down a sphere, as in the exercise given, the mechanical energy at the starting point (at rest at the top) consists entirely of potential energy, as the kinetic energy is zero. As the ball rolls down, potential energy is converted into kinetic energy maintaining the total mechanical energy unchanged. Remember, the form of energy may change, but the total amount stays the same if there are no external forces at play. Using the conservation principle allows us to set up an equation that relates the ball's potential energy at the top with its kinetic energy at a point defined by the angle \( \theta \).

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. For objects near Earth's surface, GPE is given by \( mgh \), where \( m \) is mass, \( g \) is acceleration due to gravity, and \( h \) is the height above the reference point.

In the exercise, the height \( h \) changes as the ball rolls down the sphere, and it's determined using the geometry of the situation. The key is understanding that \( h \) isn't just the ball's vertical distance from the ground; it includes the radius of the ball \( a \) and the radius of the sphere \( b \) minus any displacement caused by the angle \( \theta \).

Calculating GPE with Geometry

By applying trigonometric identities, we can link the angle \( \theta \) to the height and express GPE in terms of these variables, arriving at a mathematically precise expression for GPE as \( mgb\cos(\theta) - mga \), which is key for solving the problem and understanding how GPE changes with the motion of the ball.

Kinetic Energy of a Rolling Object

Kinetic energy (KE) is the energy of motion. For rolling objects, KE has two components: translational, due to the object moving through space, and rotational, because of the object spinning around an axis.

For an object rolling without slipping, there's a direct relationship between the linear and angular speeds: \( v = wR \). This means that for every rotation, the object travels a distance equal to the circumference of the circle formed by the rolling path. This relationship is key in finding the translational kinetic energy \( \frac{1}{2}mv^2 \) and the rotational kinetic energy \( \frac{1}{2}Iw^2 \).

Applying to the Sphere Problem

In our exercise, the ball rolls down a fixed sphere. Given the moment of inertia for a uniform ball \( I = \frac{2}{5}mR^2 \), and combining the translational and rotational kinetic energy, we get a total kinetic energy \( K = \frac{7}{10}mv^2 \) for the rolling ball. This formula allows us to understand and calculate the energy distribution in a rolling object and, together with the energy conservation law, solve for the ball’s speed in terms of \( \theta \).