Conservation of Energy with Rotation: Videos & Practice Problems

In solving rotation problems using conservation of energy, remember that kinetic energy includes both linear and rotational components. The conservation of energy equation is expressed as Ki + Ui + W__nc = Kf + Uf. When variables like velocity (v) and angular velocity (ω) appear, convert one into the other using v = r ω. This simplification aids in solving for unknowns, such as final angular speed (ω_f).

Conservation of Energy with Rotation

In solving rotation problems using the conservation of energy, it's essential to recognize that kinetic energy can be both linear and rotational. The conservation of energy equation can be expressed as:

$$K_{\text{initial}} + U_{\text{initial}} + W_{\text{nonconservative}} = K_{\text{final}} + U_{\text{final}}$$

Here, \(W_{\text{nonconservative}}\) includes work done by external forces and friction. When expanding the equation, kinetic energy \(K\) is represented as:

$$K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

Where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. For a solid disc, the moment of inertia is given by:

$$I = \frac{1}{2}mr^2$$

In problems involving both linear velocity \(v\) and angular velocity \(\omega\), it is crucial to relate these two quantities. The relationship can be expressed as:

$$v = r\omega$$

By substituting one variable for the other, you can simplify the problem to a single variable, making it easier to solve.

Consider a solid disc with mass \(m = 5 \, \text{kg}\) and radius \(r = 6 \, \text{m}\) that is free to rotate about a fixed axis. If the disc starts from rest (\(\omega_{\text{initial}} = 0\)) and a light cable is unwound from it by pulling with a constant force of \(10 \, \text{N}\) over a distance of \(8 \, \text{m}\), we can analyze the energy changes involved.

Since the disc is initially at rest, the initial kinetic energy is zero, and there is no change in potential energy. The work done by the pulling force can be calculated as:

$$W = F \cdot d \cdot \cos(\theta)$$

In this case, \(\theta = 0\) degrees, so:

$$W = 10 \, \text{N} \cdot 8 \, \text{m} \cdot 1 = 80 \, \text{J}$$

At the end of the process, the work done translates into rotational kinetic energy:

$$80 = \frac{1}{2}I\omega_{\text{final}}^2$$

Substituting the moment of inertia for the solid disc:

$$80 = \frac{1}{2} \left(\frac{1}{2} \cdot 5 \cdot 6^2\right) \omega_{\text{final}}^2$$

Solving for \(\omega_{\text{final}}\), we find:

$$80 = \frac{1}{2} \cdot \frac{1}{2} \cdot 5 \cdot 36 \cdot \omega_{\text{final}}^2$$

$$80 = 45 \omega_{\text{final}}^2$$

Thus, we can isolate \(\omega_{\text{final}}\):

$$\omega_{\text{final}} = \sqrt{\frac{80}{45}} \approx 1.33 \, \text{radians/second}$$

This example illustrates the application of conservation of energy in rotational motion, emphasizing the importance of relating linear and angular quantities to simplify calculations.

Work to accelerate cylinder

To determine the work required to accelerate a solid cylinder, we start by calculating its moment of inertia. For a solid cylinder, the moment of inertia \( I \) is given by the formula:

\[ I = \frac{1}{2} m r^2 \]

In this case, the mass \( m \) is 10 kg and the radius \( r \) is 2 m. Plugging in these values, we find:

\[ I = \frac{1}{2} \times 10 \, \text{kg} \times (2 \, \text{m})^2 = 20 \, \text{kg} \cdot \text{m}^2 \]

The cylinder is mounted to rotate freely about a perpendicular axis through its center. Initially, the cylinder is at rest, meaning its initial angular velocity \( \omega_{\text{initial}} \) is 0. We want to find the work done to accelerate it to an angular velocity of 120 RPM. To use this in calculations, we convert RPM to radians per second:

\[ \omega_{\text{final}} = 2 \pi \left(\frac{\text{RPM}}{60}\right) \]

Substituting 120 RPM gives:

\[ \omega_{\text{final}} = 2 \pi \left(\frac{120}{60}\right) = 4 \pi \, \text{radians/second} \]

Next, we apply the conservation of energy principle, which states:

\[ K_{\text{initial}} + U_{\text{initial}} + W_{\text{nonconservative}} = K_{\text{final}} + U_{\text{final}} \]

Since the cylinder starts from rest, the initial kinetic energy \( K_{\text{initial}} \) is 0, and the potential energy remains constant, allowing us to simplify the equation to:

\[ W_{\text{nonconservative}} = K_{\text{final}} \]

The final kinetic energy for rotational motion is given by:

\[ K_{\text{final}} = \frac{1}{2} I \omega^2 \]

Substituting the values we have:

\[ K_{\text{final}} = \frac{1}{2} \times 20 \, \text{kg} \cdot \text{m}^2 \times (4 \pi \, \text{radians/second})^2 \]

Calculating this yields:

\[ K_{\text{final}} = \frac{1}{2} \times 20 \times 16 \pi^2 = 160 \pi^2 \, \text{Joules} \approx 1580 \, \text{Joules} \]

Thus, the work required to accelerate the solid cylinder from rest to a speed of 120 RPM is approximately 1580 joules. This straightforward application of energy conservation illustrates the relationship between work and kinetic energy in rotational motion.

How much work is needed to stop a hollow sphere of mass 2 kg and radius 3 m that spins at 40 rad/s around an axis through its center?

• A. +9600 J
• B. −9600 J
• C. +5760 J
• D. −5760 J

Answer: (B)

Which shape reaches bottom first?

In the study of rotational motion, the concept of moment of inertia plays a crucial role in determining how different shapes of objects behave when rolling down an inclined plane. Moment of inertia, denoted as \( I \), is a measure of an object's resistance to changes in its rotational motion. It is influenced by both the mass of the object and how that mass is distributed relative to the axis of rotation. The general formula for moment of inertia is often expressed as a fraction of the form \( I = k \cdot m \cdot r^2 \), where \( k \) is a shape-dependent constant, \( m \) is the mass, and \( r \) is the radius.

For three objects of equal mass and radius but different shapes—a solid cylinder, a hollow cylinder, and a solid sphere—their moments of inertia are as follows:

• Solid Cylinder: \( I = \frac{1}{2} m r^2 \) (or 0.5 in decimal form)
• Hollow Cylinder: \( I = m r^2 \) (or 1.0 in decimal form)
• Solid Sphere: \( I = \frac{2}{5} m r^2 \) (or 0.4 in decimal form)

When these objects are released from rest at the same height on an inclined plane, their acceleration and the time taken to reach the bottom depend on their moment of inertia. The object with the lowest moment of inertia will reach the bottom first, as it experiences less rotational resistance. In this case, the solid sphere, with the lowest moment of inertia (0.4), will reach the bottom first, followed by the solid cylinder (0.5), and lastly, the hollow cylinder (1.0).

This outcome can be attributed to the distribution of mass within each shape. The solid sphere has the most optimal mass distribution, allowing it to roll down the incline with the least resistance. In contrast, the hollow cylinder has its mass concentrated at the edge, resulting in a higher moment of inertia and slower acceleration. Thus, understanding the relationship between shape, moment of inertia, and rotational motion is essential for predicting the behavior of rolling objects.

Cylinders racing down:rolling vs. sliding

In this scenario, we analyze the motion of two identical cylinders, both released from the same height but with different types of motion: one rolls without slipping while the other slides without rolling. This distinction is crucial in understanding their final speeds at the bottom of the hills.

For the cylinder that rolls without slipping, it experiences both translational and rotational motion. The relationship between the linear velocity \( v \) of the center of mass and the angular velocity \( \omega \) is given by the equation:

\( v_{\text{cm}} = r \omega \)

Here, \( r \) represents the radius of the cylinder. In contrast, the sliding cylinder has no rotational motion, meaning it only has translational velocity as it descends the hill.

The key to solving this problem lies in the principle of conservation of energy, which states that the total mechanical energy remains constant in the absence of non-conservative forces. The conservation of energy equation can be expressed as:

\( K_{\text{initial}} + U_{\text{initial}} + W_{\text{non-conservative}} = K_{\text{final}} + U_{\text{final}} \)

Initially, both cylinders have potential energy due to their height, and their kinetic energy is zero since they start from rest. As they descend, the potential energy is converted into kinetic energy. For the rolling cylinder, this kinetic energy is divided into translational kinetic energy \( K_{\text{linear}} \) and rotational kinetic energy \( K_{\text{rotational}} \). Conversely, the sliding cylinder converts all its potential energy into translational kinetic energy.

Assuming both cylinders start with the same potential energy, say 100 joules, the rolling cylinder might distribute this energy as follows: 80 joules into translational kinetic energy and 20 joules into rotational kinetic energy. The sliding cylinder, however, would convert all 100 joules into translational kinetic energy. Thus, the final kinetic energy of the sliding cylinder is greater than that of the rolling cylinder, leading to a higher final speed.

This outcome can be understood by considering energy allocation: the rolling motion requires some energy to maintain rotation, which reduces the amount of energy available for linear motion. Therefore, the sliding cylinder reaches the bottom with a greater speed due to its ability to convert all potential energy into translational kinetic energy, while the rolling cylinder splits its energy between two forms of kinetic energy.

In summary, the cylinder that slides down the hill achieves a higher final speed than the one that rolls, illustrating the impact of energy distribution in different types of motion.

Two solid cylinders of same mass and radius roll on a horizontal surface just before going up an inclined plane. Cylinder A rolls without slipping, but cylinder B moves along a slippery path, so it moves without rotating at all times. At the bottom of the incline, both have the same speed at their center of mass. Which will go higher on the inclined plane? (Why?)

• A. Cylinder A
• B. Cylinder B
• C. Both reach the same height

Answer: (A)

Here’s what students ask on this topic:

What is the conservation of energy equation for rotational motion?

The conservation of energy equation for rotational motion is similar to that for linear motion but includes rotational kinetic energy. It is expressed as:

$K_i+U_i+W_{\mathrm{nc}}=K_f+U_f$

Here, $K_i$ and $K_f$ are the initial and final kinetic energies, which can include both linear and rotational components. $U_i$ and $U_f$ are the initial and final potential energies, and $W_{\mathrm{nc}}$ is the work done by non-conservative forces.

How do you convert linear velocity to angular velocity in rotational motion problems?

To convert linear velocity ($v$) to angular velocity ($\omega$), you use the relationship:

$v=r\omega$

Here, $r$ is the radius of the circular path. This equation allows you to express one variable in terms of the other, simplifying the problem to a single variable.

What is the moment of inertia for a solid disc?

The moment of inertia ($I$) for a solid disc rotating about an axis through its center is given by:

$I=\frac{1}{2}m{r}^2$

Here, $m$ is the mass of the disc, and $r$ is its radius.

How do you solve for the final angular speed using conservation of energy?

To solve for the final angular speed ($\omega$_f) using conservation of energy, follow these steps:

1. Write the conservation of energy equation: $K_i+U_i+W_{\mathrm{nc}}=K_f+U_f$

2. Expand the kinetic and potential energy terms. For rotational motion, kinetic energy is $\frac{1}{2}I{\omega }^2$.

3. Substitute known values and solve for $\omega$_f.

Example: If a disc with moment of inertia $I$ is acted upon by a force doing work $W$, the final angular speed is found by:

$W=\frac{1}{2}I{\omega }^2$

Solve for $\omega$:

$\omega =\sqrt{\frac{2W}{I}}$

What is the significance of 'without slipping' in rotational motion problems?

The term 'without slipping' in rotational motion problems indicates that the point of contact between the rotating object and the surface does not slide. This condition allows us to use the relationship:

$v=r\omega$

Here, $v$ is the linear velocity, $r$ is the radius, and $\omega$ is the angular velocity. This relationship simplifies the problem by linking linear and rotational motion, making it easier to solve for unknowns.