Problem 16 - Classical Mechanics - An Undergraduate Text

Question

A uniform ball of radius \(a\) and centre \(G\) can roll without slipping on the inside surface of a fixed hollow sphere of (inner) radius \(b\) and centre \(O\). The ball undergoes planar motion in a vertical plane through \(O\). Find the energy conservation equation for the ball in terms of the variable \(\theta\), the angle between the line \(O G\) and the downward vertical. Deduce the period of small oscillations of the ball about the equilibrium position.

Step-by-Step Solution

Answer

The energy conservation equation is given by \(7gb = 7ga\cos\(\theta\) + \frac{7}{10}v^2\), and the period of small oscillations of the ball about the equilibrium position is \(T = 2\pi\sqrt{\frac{b}{g}}\)

Step 1: Setup the problem

The ball is moving inside a hollow sphere. Let's denote \(m\) as the mass of the ball. We can break this motion down into two parts: translational and rotational motion. The distance from the centre \(O\) of the sphere to the centre \(G\) of the ball is \(b-a\cos\(\theta\)\). The height of the center \(G\) from the bottom of the sphere is \(b-a\cos\(\theta\)\)

Step 2: Calculate the Potential Energy (PE)

The potential energy is given by \(PE = mgh\), where \(g\) is the acceleration due to gravity and \(h\) is the height of the object. Substituting the height \(h = b - a\cos\(\theta\)\), we get \(PE = mgb - mga\cos\(\theta\)\)

Step 3: Calculate the Kinetic Energy (KE)

The kinetic energy can be expressed in terms of translational energy and rotational energy. The translational kinetic energy is \(\frac{1}{2}mv^2\) and the rotational kinetic energy of a rolling ball is \(\frac{1}{2}I\omega^2\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. The moment of inertia \(I\) for a uniform ball is \(\frac{2}{5}mr^2\), and because the ball is rolling without slipping, \(\omega = \frac{v}{r}\), where \(r\) is the radius of the ball. Substituting these in we get \(KE = \frac{7}{10}mv^2\)

Step 4: Apply conservation of Energy

Since no external work is done on the ball, the total mechanical energy should be conserved. This implies that the total initial energy should equal to total final energy. Hence \(PE_i + KE_i = PE_f + KE_f = Constant\). This means \(mgb + 0 = mgb - mga\cos\(\theta\) + \frac{7}{10}mv^2\), then simplify to get \(7gb = 7ga\cos\(\theta\) + \frac{7}{10}v^2\)

Step 5: Finding the equation for \(\theta\)

Now we differentiate \(7gb = 7ga\cos\(\theta\) + \frac{7}{10}v^2\) equation with respect to time, and solve for \(\theta\) with the small angle approximation.

Step 6: Calculate the period of small oscillations

To calculate the period \(T\) of small oscillations, we can use the formula \(T = 2\pi\sqrt{\frac{l}{g}}\), where \(l\) is the effective length of the pendulum. This results in \(T = 2\pi\sqrt{\frac{b}{g}}\)

Key Concepts

Potential Energy

Potential energy is the energy stored in an object due to its position, and for the ball rolling inside the hollow sphere, it depends on its height above the lowest point of the sphere. When the ball is at the lowest point, its potential energy is zero, but as it moves up, the energy increases proportionally to its height.
The formula for potential energy is given by \(PE = mgh\), where \(m\) is the mass of the ball, \(g\) is the acceleration due to gravity, and \(h\) is the height above the reference point.
In the exercise, the height \(h\) is given by \(b-a\cos(\theta)\), where \(b\) is the radius of the sphere, and \(a\) is the radius of the ball.

• Thus, the potential energy of the ball is \(PE = mg(b - a\cos(\theta))\).
• This formula helps us understand how potential energy changes as the ball moves, which is crucial for applying the principle of conservation of energy.

Through this relationship, energy conservation becomes evident: as the height (and thus potential energy) changes, kinetic energy adjusts to balance the total energy.

Kinetic Energy

Kinetic energy is the energy of motion. For the rolling ball, it comprises both translational and rotational kinetic energy. This dual nature occurs because the ball not only moves in a straight line but also rotates around its own axis.
The translational kinetic energy is given by \(\frac{1}{2}mv^2\), where \(v\) is the velocity of the center of mass of the ball.
The rotational kinetic energy depends on the ball's moment of inertia \(I\) and angular velocity \(\omega\). The moment of inertia for a uniform ball is \(\frac{2}{5}mr^2\). Since the ball rolls without slipping, \(\omega\) relates to \(v\) through \(\omega = \frac{v}{r}\).

• Combining these, the kinetic energy of the ball is \(KE = \frac{7}{10}mv^2\).
• This reflects the total motion energy, capturing the ball's speed as well as its spinning action.

This kinetic energy is a key component when applying the conservation of energy principle to determine how energy is distributed between potential and kinetic forms as the ball oscillates inside the sphere.

Period of Small Oscillations

The period of small oscillations refers to the time it takes for the ball to complete one full back-and-forth swing around its equilibrium position. It is closely related to the concepts of harmonic motion and involves analyzing the forces and energies at play when the ball is slightly displaced from equilibrium.
In this setup, the ball's motion can resemble a simple pendulum when oscillations are small.
We derive the period from the formula for a simple pendulum: \(T = 2\pi\sqrt{\frac{l}{g}}\), where \(l\) is the effective length, which, in this context, is the radius of the hollow sphere \(b\).

• Thus, for the ball, when it undergoes small oscillations, the period \(T = 2\pi\sqrt{\frac{b}{g}}\).
• This equation allows us to predict how fast the ball will oscillate based on the size of the hollow sphere and the acceleration due to gravity.

This understanding helps in determining dynamics such as frequency and timing of oscillations, essential components in more complex mechanics studies.