Problem 61 - University Physics with Modern Physics

Question

CALC Rotating Bar. A thin, uniform 12.0 -kg bar that is 2.00 \(\mathrm{m}\) long rotates uniformly about a pivot at one end, making 5.00 complete revolutions every 3.00 seconds. What is the kinetic energy of this bar? (Hint. Different points in the bar have different speeds. Break the bar up into infinitesimal segments of mass dm and integrate to add up the kinetic energies of all these segments.)

Step-by-Step Solution

Answer

The kinetic energy of the rotating bar is approximately 438.2 J.

Step 1: Define the Problem

We need to find the kinetic energy of a rotating bar. The hint suggests breaking the bar into infinitesimal segments and calculating their kinetic energy individually before integrating over the entire bar.

Step 2: Express Segment Properties

Consider an infinitesimal segment of the bar at a distance \( x \) from the pivot with a thickness \( dx \). The mass of this segment is given by \( dm = \frac{M}{L} dx \), where \( M = 12.0 \, \text{kg} \) and \( L = 2.0 \, \text{m} \) are the mass and length of the bar, respectively.

Step 3: Express Segment Velocity

The linear velocity \( v \) of the segment is dependent on its distance \( x \) from the pivot: \( v = \omega x \), where \( \omega \) is the angular velocity. We need to find \( \omega \) in rad/s given that the bar makes 5.00 revolutions every 3.00 seconds.

Step 4: Calculate Angular Velocity

One revolution is \( 2\pi \) radians. Therefore, \( \omega = \frac{5 \times 2\pi}{3} \, \text{rad/s} \approx 10.47 \, \text{rad/s} \).

Step 5: Express Kinetic Energy of Segment

The kinetic energy \( dK \) of the segment is \( dK = \frac{1}{2} dm \cdot v^2 = \frac{1}{2} \left( \frac{M}{L} dx \right) (\omega x)^2 \).

Step 6: Integrate to Find Total Kinetic Energy

Integrate \( dK \) from \( x = 0 \) to \( x = L \) to find total kinetic energy \( K \): \[ K = \int_{0}^{L} \frac{1}{2} \frac{M}{L} \omega^2 x^2 \, dx = \frac{1}{2} \frac{M}{L} \omega^2 \int_{0}^{L} x^2 \, dx. \]

Step 7: Solve the Integral

Calculate the integral: \[ \int_{0}^{L} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{L} = \frac{L^3}{3}. \]Substitute this into the kinetic energy expression: \[ K = \frac{1}{2} \frac{M}{L} \omega^2 \times \frac{L^3}{3} = \frac{1}{6} M \omega^2 L^2. \]

Step 8: Calculate Final Kinetic Energy

Using \( M = 12.0 \, \text{kg} \), \( L = 2.0 \, \text{m} \), and \( \omega \approx 10.47 \, \text{rad/s} \), calculate:\[ K = \frac{1}{6} (12) (10.47)^2 (2)^2 \approx 438.2 \, \text{J}. \]

Key Concepts

Kinetic Energy

Kinetic energy is a fundamental concept in physics that describes the energy an object possesses due to its motion. For rotational motion, kinetic energy depends not only on the mass of the object but also on how that mass is distributed relative to the axis of rotation. The formula to calculate the kinetic energy for a rotating body is:\[ K = \frac{1}{2} I \omega^2 \]where:

• \( I \) is the moment of inertia, which quantifies the distribution of mass around the pivot point.
• \( \omega \) is the angular velocity.

In this specific problem, the bar is broken into tiny segments. Each segment contributes to the total kinetic energy.

The integration over these infinitesimal segments accounts for changing speeds along the bar, ensuring that the kinetic energy calculation is precise. This involves considering that kinetic energy also depends on the velocity, which varies for each segment. Understanding kinetic energy in rotational motion allows us to appreciate how distribution and rotation affect the energy.

Angular Velocity

Angular velocity, symbolized as \( \omega \), refers to how fast an object rotates or spins. Just like linear velocity tells us how fast an object moves across a distance in a given time, angular velocity provides the rate of rotation. It's generally measured in radians per second (rad/s).

In the problem scenario, the bar completes 5 revolutions in 3 seconds. To find the angular velocity, we first convert revolutions into radians:

• 1 revolution = \( 2\pi \) radians
• So, 5 revolutions = \( 5 \times 2\pi \) radians

Divide this angular distance by the time it takes to complete, which is 3 seconds:\[\omega = \frac{5 \times 2\pi}{3}\]
This yields an approximate value for \( \omega \) of 10.47 rad/s. Understanding angular velocity is crucial when analyzing any rotating system, as it directly influences the kinetic energy of the system.

Integral Calculus

Integral calculus is a branch of mathematics used to find quantities like areas, volumes, and in this case, the total kinetic energy by summing infinitesimal differences. This process is essential when dealing with continuous distributions of mass, such as a rotating bar.

In this exercise, we use integral calculus to add up the kinetic energy contributions from infinitesimal segments of the bar. To find the total kinetic energy:\[ K = \int_{0}^{L} \frac{1}{2} \frac{M}{L} \omega^2 x^2 \, dx \]Here’s the breakdown:

• \( x \) represents the position of a segment from the pivot.
• \( dx \) is the infinitesimal width of each segment.
• The integral adds the energies of all these segments from \( x=0 \) (pivot) to \( x=L \) (end of bar).

Solving this integral requires applying the power rule, which here gives:\[ \int_{0}^{L} x^2 \, dx = \frac{L^3}{3} \]Thus, integral calculus allows us to transition from a sum over smaller pieces to a complete understanding of the kinetic energy in terms of a single expression. This approach is critical in physics for accuracy in non-uniform distributions.