Question

\({ }^{\dagger} \mathrm{A}\) thin uniform disc, mass \(M\), radius \(a\) and centre \(C\), has a thin uniform rod \(O C\), mass \(m\) and length \(a \sqrt{3}\), fixed to it at \(C\), so that \(O C\) is orthogonal to the disc. The end \(O\) of the rod is fixed but freely pivoted at the centre \(O\) of a horizontal turntable, and the rim of the disc rests on the surface of the turntable. No slipping occurs. The turntable is forced to rotate about the vertical axis through \(O\) with variable angular velocity \(\Omega\). Initially the system is at rest. Show that if \(P\) is the point of contact between the disc and the turntable, and \(\varphi\) is the angle between \(O P\) and a line fixed in the turntable, then $$ \dot{\varphi}=-\left(\frac{11 M+4 m}{19 M+4 m}\right) \Omega $$

Short answer

Answer: $$\dot{\varphi} = -(\frac{11M+4m}{19M+4m})\Omega$$

Step-by-step solution

Find the moment of inertia of the system

The moment of inertia of the disc (with mass \(M\) and radius \(a\)) about axis \(OC\) can be found using the perpendicular axis theorem, which gives \(I_{disc} = \frac{1}{2}Ma^2\). The moment of inertia of the rod (with mass \(m\) and length \(a\sqrt{3}\)) about axis \(OC\) is \(I_{rod} = \frac{1}{3}ma^2\). The total moment of inertia of the system about axis \(OC\) is the sum of the moment of inertia of the disc and rod: \(I_{total} = I_{disc} + I_{rod} = \frac{1}{2}Ma^2 + \frac{1}{3}ma^2\).

Write the conservation of angular momentum equation

The conservation of angular momentum states that the initial angular momentum equals the final angular momentum. Since the initial angular momentum is zero (the system starts at rest), the final angular momentum must also be zero. This can be written as: \(I_{total} \Omega_f + m\dot{\varphi} a^2 = 0\), where \(\Omega_f\) is the final angular velocity after some time and \(\dot{\varphi}\) is the rate of change of angle \(\varphi\).

Solve the conservation of angular momentum equation for \(\dot{\varphi}\)

We now solve the conservation of angular momentum equation for \(\dot{\varphi}\). Using the expression for the total moment of inertia, we have \((\frac{1}{2}Ma^2 + \frac{1}{3}ma^2) \Omega_f + ma^2\dot{\varphi} = 0\). Rearrange the equation to isolate \(\dot{\varphi}\) on one side: \(\dot{\varphi} = -(\frac{11M+4m}{19M+4m})\Omega_f\). Since \(\Omega_f\) is the final angular velocity, it must be equal to \(\Omega\), therefore: $$\dot{\varphi} = -(\frac{11M+4m}{19M+4m})\Omega$$

Key concepts

Moment of Inertia

At the heart of rotational dynamics is the concept of moment of inertia, often signified by the symbol 'I'. Moment of inertia is the rotational equivalent of mass in linear motion. It represents an object's resistance to changes in its rotation rate. Imagine you're trying to spin a wheel. How easily it starts spinning depends on its moment of inertia. Unlike mass, which depends solely on the quantity of matter, moment of inertia depends not only on the mass of an object but also on how that mass is distributed in relation to the rotational axis.

Moment of inertia is calculated by summing the products of the mass of each particle in the body and the square of its distance from the rotational axis. For uniform, symmetrical objects, there are standard formulas. For example, the moment of inertia of a thin disc about its axis is given by \( I_{disc} = \frac{1}{2}M a^2 \) where 'M' is the mass and 'a' is the radius. For a rod of length \( l \) and mass \( m \) pivoted at the end, the moment of inertia is \( I_{rod} = \frac{1}{3}ml^2 \) provided the rod swings in a plane perpendicular to its length. When combined in a system, like a disc and rod in our exercise, their individual moments of inertia add up to get the system's overall moment of inertia.

Conservation of Angular Momentum

Now let's delve into the principle of conservation of angular momentum. This principle tells us that if no external torque acts on a system, its total angular momentum remains constant. Angular momentum, analogous to linear momentum in non-rotating systems, is determined by the product of the object's moment of inertia and its angular velocity \( \Omega \).

In the absence of external forces, such as friction with the air or a force at the pivot point, our system's angular momentum initially is zero - because it starts at rest. As the system begins to rotate due to the turntable's movement, the conservation of angular momentum means that the angular momentum of one part of the system must be balanced by an opposite change in another - leading to the derived equation \( I_{total} \Omega_f + m \dot{\varphi} a^2 = 0 \), where \( I_{total} \) is the total moment of inertia and \( \dot{\varphi} \) is the rate of change of the angle \( \varphi \) as observed in the exercise. By understanding and applying this principle, we can solve complex problems pertaining to rotational motion and predict the behavior of rotating systems.

Rotational Motion

Lastly, we explore the broader concept of rotational motion which underlies our exercise. Rotational motion is movement around an axis. This can become complicated because different parts of an object can have different velocities and accelerations even though they are rigidly connected. Think of a spinning figure skater: as they pull in their arms, they spin faster. But why?

During rotational motion, all points in an object move in circles around the rotation axis. These circles have radii that are equal to the distance from the axis. Because the circumference of a circle is \( 2\pi r \) (where 'r' is the radius), points farther from the axis have to travel a greater distance in the same amount of time. Therefore, they move faster, meaning they have greater linear velocity even though they have the same angular velocity as points closer to the axis.

In our exercise, as the turntable spins, its point of contact with the disc induces a rotational motion in the disc. This interaction is intricate because the disc and the attached rod are subjected to different forces and have different moments of inertia. By tying together the moment of inertia and the conservation of angular momentum, we can describe the complex motion observed in the system with equations like \( \dot{\varphi} = -\left(\frac{11M+4m}{19M+4m}\right)\Omega \) where \( \dot{\varphi} \) represents the angular velocity at which the line OP rotates relative to the turntable.