# Energy in an angular (coil) spring system

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In summary: Thanks Berkeman! Yes, then there would be no energy loss and the amplitude should be the same. However, after thinking it through I have arrived at some other calculations. Firstly since angular momentum is conserved and the angular (radian) speed (what is it called now again?) in fact is given by omega*theta this should lead to I_{0}\omega_{0}\theta_{0}=3I_{0}\frac{\omega_{0}}{\sqrt{3}}\theta_{new}, which leads to \theta_{new}=\frac{\theta_{0}}{\sqrt{3}}. Secondly, since the physical omega comes into the energy equations this should lead
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## Homework Statement

A solid disk of mass M and radius R is on a vertical shaft. The shaft is attached to a coil spring which exerts a linear restoring torque of magnitude $$C\theta,$$ where theta is the angle measured from the static equilibrium position and C is constant. Neglect the mass of the shaft and the spring, and assume the bearings to be frictionless. (Please tell me if you want me to attach a figure as well. That is a lot of work but can be arranged.)

a. Show that the disc can undergo simple harmonic motion, and find the frequency of the motion.

b. Suppose that the disc is moving according to $$\theta=\theta_{0}\sin(\omega t),$$ where omega is the frequency found in part a. At time $$t_{1}=\pi / \omega ,$$ a ring of sticky putty of mass M and radius R is dropped concentrically on the disk. Find:
(1) The new frequency of the motion
(2) The new amplitude of the motion

## Homework Equations

$$I_{disc}=\frac{MR^{2}}{2}$$
$$I_{ring}=MR^{2}$$

## The Attempt at a Solution

a. The initial moment of inertia is $$I_{0}=\frac{MR^{2}}{2}$$ and the torque is defined as $$\tau = I \ddot{\theta}.$$ In this case the torque is $$\tau=-C\theta,$$ which leads to the equation $$\ddot{\theta}=-\frac{2C}{MR^{2}}\theta,$$ with the solution $$\theta=\theta_{0}\sin (\omega t),$$ where $$\omega=\sqrt{\frac{2C}{MR^{2}}}.$$

b. Differentiating the equation leads to $$\dot{\theta}=\theta_{0}\omega\cos (\omega t),$$ so $$\dot{\theta}(\pi / \omega)=-\theta_{0} \omega,$$ so the disc has a certain velocity. Calculating the new frequency is easy: $$\omega_{new}=\sqrt{\frac{C}{\frac{MR^{2}}{2}+MR^{2}}}=\sqrt{\frac{2C}{3MR^{2}}}$$ Now since the angular momentum is conserved $$\theta_{0}\omega=\theta_{new}\omega_{new},$$ so $$\theta_{new}=\theta_{0}\sqrt{3}.$$

Now can this be correct? The energy is defined as $$\frac{I\omega^{2}}{2}.$$ So in this case $$E_{i}=\frac{I_{0}\omega^{2}}{2}=C$$ and $$E_{f}=\frac{(I_{0}+I_{ring})\omega_{new}^{2}}{2}=C$$ so the energy is conserved. I don't understand this because there is a collision and some energy should be lost! Does this mean that my calculation is incorrect. And if so, where did I go wrong?

Order said:

## Homework Statement

A solid disk of mass M and radius R is on a vertical shaft. The shaft is attached to a coil spring which exerts a linear restoring torque of magnitude $$C\theta,$$ where theta is the angle measured from the static equilibrium position and C is constant. Neglect the mass of the shaft and the spring, and assume the bearings to be frictionless. (Please tell me if you want me to attach a figure as well. That is a lot of work but can be arranged.)

a. Show that the disc can undergo simple harmonic motion, and find the frequency of the motion.

b. Suppose that the disc is moving according to $$\theta=\theta_{0}\sin(\omega t),$$ where omega is the frequency found in part a. At time $$t_{1}=\pi / \omega ,$$ a ring of sticky putty of mass M and radius R is dropped concentrically on the disk. Find:
(1) The new frequency of the motion
(2) The new amplitude of the motion

## Homework Equations

$$I_{disc}=\frac{MR^{2}}{2}$$
$$I_{ring}=MR^{2}$$

## The Attempt at a Solution

a. The initial moment of inertia is $$I_{0}=\frac{MR^{2}}{2}$$ and the torque is defined as $$\tau = I \ddot{\theta}.$$ In this case the torque is $$\tau=-C\theta,$$ which leads to the equation $$\ddot{\theta}=-\frac{2C}{MR^{2}}\theta,$$ with the solution $$\theta=\theta_{0}\sin (\omega t),$$ where $$\omega=\sqrt{\frac{2C}{MR^{2}}}.$$

b. Differentiating the equation leads to $$\dot{\theta}=\theta_{0}\omega\cos (\omega t),$$ so $$\dot{\theta}(\pi / \omega)=-\theta_{0} \omega,$$ so the disc has a certain velocity. Calculating the new frequency is easy: $$\omega_{new}=\sqrt{\frac{C}{\frac{MR^{2}}{2}+MR^{2}}}=\sqrt{\frac{2C}{3MR^{2}}}$$ Now since the angular momentum is conserved $$\theta_{0}\omega=\theta_{new}\omega_{new},$$ so $$\theta_{new}=\theta_{0}\sqrt{3}.$$

Now can this be correct? The energy is defined as $$\frac{I\omega^{2}}{2}.$$ So in this case $$E_{i}=\frac{I_{0}\omega^{2}}{2}=C$$ and $$E_{f}=\frac{(I_{0}+I_{ring})\omega_{new}^{2}}{2}=C$$ so the energy is conserved. I don't understand this because there is a collision and some energy should be lost! Does this mean that my calculation is incorrect. And if so, where did I go wrong?

Nice work. Consider if the putty is dropped on the disk at the moment when it is at the full travel in one direction (so it is momentarily not moving). There is no "collision" in the direction of the oscillatory motion, just a change in the moment of inertia...

berkeman said:
Nice work. Consider if the putty is dropped on the disk at the moment when it is at the full travel in one direction (so it is momentarily not moving). There is no "collision" in the direction of the oscillatory motion, just a change in the moment of inertia...

Thanks Berkeman! Yes, then there would be no energy loss and the amplitude should be the same. However, after thinking it through I have arrived at some other calculations. Firstly since angular momentum is conserved and the angular (radian) speed (what is it called now again?) in fact is given by omega*theta this should lead to $$I_{0}\omega_{0}\theta_{0}=3I_{0}\frac{\omega_{0}}{\sqrt{3}}\theta_{new},$$ which leads to $$\theta_{new}=\frac{\theta_{0}}{\sqrt{3}}.$$
And since the physical omega comes into the energy equations this should lead to that the energy is not the spring constant, but the spring constant times the amplitude squared, i. e. $$E=\frac{C\theta^{2}}{2}$$ and therefore there is a loss of 2/3 of the energy. If I put it on the disk "when it is at the full travel in one direction" the amplitude will not change and there will be no energy loss at all.

Now if this is wrong I don't know what to do. I can't get any further.

## 1. What is an angular (coil) spring system?

An angular (coil) spring system is a mechanical system that consists of a coiled spring attached to a rotating object, such as a wheel or a pendulum. The spring stores potential energy when it is compressed or stretched, and this energy is released as the object rotates.

## 2. How does energy transfer occur in an angular (coil) spring system?

Energy transfer occurs in an angular (coil) spring system through the conversion of potential energy into kinetic energy. When the spring is compressed or stretched, it gains potential energy. As the rotating object begins to move, the potential energy is converted into kinetic energy, causing the object to rotate.

## 3. What factors affect the amount of energy stored in an angular (coil) spring system?

The amount of energy stored in an angular (coil) spring system is affected by several factors, including the stiffness of the spring, the amount of compression or stretching, and the mass of the rotating object. The more the spring is compressed or stretched, the more potential energy is stored, and a stiffer spring will store more energy than a less stiff spring.

## 4. How can the energy in an angular (coil) spring system be calculated?

The energy in an angular (coil) spring system can be calculated using the equation E = 1/2 * k * x^2, where E is the energy stored in the spring, k is the spring constant, and x is the distance the spring is compressed or stretched. The spring constant can be determined experimentally or by using the material properties of the spring.

## 5. What are some real-world applications of an angular (coil) spring system?

Angular (coil) spring systems have many real-world applications, including in watches, clocks, and other timekeeping devices, as well as in suspension systems for vehicles. They are also used in toys, exercise equipment, and various mechanical devices that require a source of stored energy and controlled release.

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