Two thin rods of length L are rotating with the same angular speed w (in rad/s) about axes that pass perpendicularly th end. Rod A is massless but has a particle of mass 0.70 kg attached to its free end. Rod B has a'mass of 0.70 kg, which is uniformly along its length. The length of each rod is 0.72 m, and the angular speed is 4.9 rad/s. Find the kinetic energie- its attached particle and of rod B. (a) KEA = i
Angular Momentum
The momentum of an object is given by multiplying its mass and velocity. Momentum is a property of any object that moves with mass. The only difference between angular momentum and linear momentum is that angular momentum deals with moving or spinning objects. A moving particle's linear momentum can be thought of as a measure of its linear motion. The force is proportional to the rate of change of linear momentum. Angular momentum is always directly proportional to mass. In rotational motion, the concept of angular momentum is often used. Since it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics. To understand the concept of angular momentum first we need to understand a rigid body and its movement, a position vector that is used to specify the position of particles in space. A rigid body possesses motion it may be linear or rotational. Rotational motion plays important role in angular momentum.
Moment of a Force
The idea of moments is an important concept in physics. It arises from the fact that distance often plays an important part in the interaction of, or in determining the impact of forces on bodies. Moments are often described by their order [first, second, or higher order] based on the power to which the distance has to be raised to understand the phenomenon. Of particular note are the second-order moment of mass (Moment of Inertia) and moments of force.
![**Rotational Kinetic Energy of Two Thin Rods**
*Problem Statement:*
Two thin rods of length \( L \) are rotating with the same angular speed \( \omega \) (in rad/s) about axes that pass perpendicularly through one end. Rod A is massless but has a particle of mass 0.70 kg attached to its free end. Rod B has a mass of 0.70 kg, which is distributed uniformly along its length. The length of each rod is 0.72 m, and the angular speed is 4.9 rad/s. Find the kinetic energies of rod A with its attached particle and of rod B.
1. (a) \( KE_A \) =
2. (b) \( KE_B \) =
### Explanation:
#### Kinetic Energy Calculation for Rod A:
Rod A is considered massless, but it has a particle of mass \( m \) attached to its free end. The kinetic energy \( KE_A \) can be calculated using the formula for rotational kinetic energy for a point mass:
\[ KE = \frac{1}{2} m L^2 \omega^2 \]
Given that:
- \( m = 0.70 \) kg
- \( L = 0.72 \) m
- \( \omega = 4.9 \) rad/s
\[ KE_A = \frac{1}{2} \times 0.70 \, \text{kg} \times (0.72 \, \text{m})^2 \times (4.9 \, \text{rad/s})^2 \]
#### Kinetic Energy Calculation for Rod B:
Rod B has its mass uniformly distributed along its length. The kinetic energy \( KE_B \) can be calculated using the formula for rotational kinetic energy of a rod rotating about one end:
\[ KE = \frac{1}{6} m L^2 \omega^2 \]
Given that:
- \( m = 0.70 \) kg
- \( L = 0.72 \) m
- \( \omega = 4.9 \) rad/s
\[ KE_B = \frac{1}{6} \times 0.70 \, \text{kg} \times (0.72 \, \text{m})^2 \times (4.9 \, \text{rad/s})^2 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F881eeb20-3739-4314-bfbb-cdc248cde7bb%2Ffc16da8a-97af-4270-af3f-ee49db1fa14b%2Fm97ewpm_processed.jpeg&w=3840&q=75)
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