Problem 14 - University Physics with Modern Physics

Question

A small cylinder of mass m can slide without friction on a shaft that is attached to a turntable, as shown in the figure. The shaft also passes through the coils of a spring with spring constant \(k\), which is attached to the turntable at one end, and to the cylinder at the other end. The equilibrium length of the spring (unstretched and uncompressed) matches the radius of the turntable; thus, when the turntable is not

Step-by-Step Solution

Answer

Question: When a turntable with a cylinder connected to a spring is set rotating with an angular speed \(\omega\), find an expression for the distance, \(r_e\), the cylinder moves along the shaft from its original position in terms of \(k\), \(m\), and \(\omega\). Answer: \(r_e = \frac{-r_0m\omega^2}{k - m\omega^2}\)

Step 1: Centripetal and spring forces

The centripetal force acting on the cylinder (keeping it moving in a circle) is given by: $$F_{c} = m\omega^2 r$$ Where: \(m\) - mass of the cylinder \(\omega\) - Angular speed of the turntable \(r\) - Radius of the circle the cylinder moves on The spring force (which opposes the centripetal force) is given by Hooke's Law: $$F_{s} = k(r_e+r_0)$$ Where: \(k\) - Spring constant \(r_e\) - Distance the cylinder moves along the shaft \(r_0\) - Unstretched (equilibrium) length of the spring (equal to radius of the turntable)

Step 2: Equating forces and solving for \(r_e\)

At equilibrium, the centripetal force and the spring force acting on the cylinder must be equal: $$m\omega^2r = k(r_e + r_0)$$ Considering that the cylinder moves on a circle with a radius equal to \(r = r_e + r_0\) (the sum of the initial radius and the additional distance): $$m\omega^2(r_e + r_0) = k(r_e + r_0)$$ The goal now is to find the expression for \(r_e\). By dividing both sides by \((r_e + r_0)\) and reorganizing the equation, we get: $$r_e(k - m\omega^2) = -r_0m\omega^2$$ Finally, solve for \(r_e\) by dividing both sides by \((k - m\omega^2)\):

Step 3: Final expression for \(r_e\)

The expression for the distance the cylinder moves along the shaft, \(r_e\), in terms of \(k\), \(m\), and \(\omega\) is: $$r_e = \frac{-r_0m\omega^2}{k - m\omega^2}$$

Key Concepts

Centripetal Force

When an object is moving in a circular path, it experiences a force directed towards the center of the circle, which is known as the centripetal force. This inward force is essential for any type of circular motion and always acts perpendicular to the object's motion, preventing it from flying outwards due to inertia.

In the context of our exercise, the centripetal force is what keeps the cylinder moving along with the turntable as it rotates. The formula for centripetal force is given by: \[\begin{equation}F_c = m\text{\(\omega^2\)}r\text{\(\end{equation}\]\)} where \(m\) is the mass of the cylinder, \(\omega\) is the angular speed of the turntable, and \(r\) is the radius of the circle the cylinder moves on.

Understanding centripetal force is critical in various fields, such as designing roller coasters or calculating the orbits of planets and satellites. In this problem, harnessing the idea of centripetal force allows us to explore the balance of forces at play and the resulting movement of objects in rotational dynamics.

Hooke's Law

Hooke's Law is a principle that describes the behavior of springs and other elastic objects. It states that the force needed to extend or compress a spring by some distance is proportional to that distance. This is mathematically expressed as: \[\begin{equation}F_s = kx\text{\(\end{equation}\]\)}where \(F_s\) is the spring force, \(k\) is the spring constant (a measure of the spring's stiffness), and \(x\) is the displacement from the spring's equilibrium position.

In our exercise, Hooke's Law is applied to determine the restoring force exerted by the spring when the cylinder is displaced by a distance \(r_e\) from its equilibrium position. The force exerted by the spring opposes the centripetal force, illustrating an equilibrium condition in rotational systems. By understanding Hooke's Law, students can solve for the extension or compression of springs in various physical contexts, from simple pendulums to complex engineering systems.

Angular Speed

The concept of angular speed is fundamental in rotational dynamics. It is defined as the rate at which an object rotates or revolves around a fixed axis, measured in radians per second. For an object moving in a circular path, angular speed can be thought of as how quickly it is moving around the circle.

In symbols, angular speed is represented by \(\omega\) and is a crucial element in the equations for centripetal force. High angular speeds result in greater centripetal forces, which has implications for the integrity of rotating systems, including engines, turbines, and even carnival rides.

In the exercise presented, angular speed determines the magnitude of the centripetal force that, in turn, affects the extent to which the spring is stretched according to Hooke's Law. Therefore, being conversant with angular speed enables one to predict and calculate the behavior of objects in circular motion, bridging the gap between theoretical physics and practical applications.