10 Fixed-Axis Rotation
10.7 Newton's Second Law for Rotation
Learning Objectives
Calculate the torques on rotating systems about a fixed axis to find the angular accelerationExplain how changes in the moment of inertia of a rotating system affect angular acceleration with a fixed applied torqueIn this section, we are integrating the concepts we have learned so far to examine the dynamics of rotating rigid bodies. While we have explored motion through kinematics and rotational kinetic energy, we have not yet linked these concepts with force or torque. Here, we introduce the rotational counterpart to Newton's second law of motion and apply it to rigid bodies undergoing fixed-axis rotation.
Newton's Second Law for Rotation
We have identified various counterparts in rotational terms to the translational concepts discussed earlier, torque being the most recent, serving as the rotational equivalent of force. This leads us to question whether there is an analogous equation to Newton's second law, ΣF = ma, involving torque and rotational motion.To investigate this, let's consider Newton's second law for a single particle rotating around an axis and moving in circular motion. Imagine applying a force on a point mass at a distance 'r' from a pivot point. The particle is constrained to move in a circular path with a fixed radius, and the force is tangent to the circle. By applying Newton's second law, we can determine the magnitude of the acceleration 'a' as F/m in the direction of the force. Remember, the tangential acceleration's magnitude is proportional to the angular acceleration's magnitude, given by a = rα. Substituting this into Newton's second law yields:\[ a = \frac{F}{m} \]In simpler terms, this equation showcases the relationship between force, mass, and acceleration in rotational motion. For instance, imagine a spinning top. When you apply a force to it, the top accelerates in the direction of the force applied. The rate at which it accelerates depends on the force exerted and the top's mass. This acceleration is crucial in understanding how objects rotate and respond to external influences.By grasping these fundamental concepts, you'll be equipped to analyze and interpret the complexities of rotational dynamics with more clarity and insight.
Summary of Physics Concepts
Newtons’s Second Law and Rotational Dynamics
- Newton’s Second Law in Rotation: Newton's Second Law can be extended to rotational motion. In rotation, the sum of torques acting on an object equals the moment of inertia times the angular acceleration.
- Torque and Moment of Inertia: Torque is the rotational equivalent of force in linear motion. It is calculated as the product of the lever arm and the force applied perpendicularly. Moment of inertia, denoted by I, is the rotational analog of mass in linear motion and is given by I = mr^2.
- Angular Acceleration: Angular acceleration represents how quickly an object's angular velocity changes over time. It is related to the torque applied and the moment of inertia of the object.
- Rotational Dynamics of Rigid Bodies: When multiple torques act on a rigid body rotating about a fixed axis, the sum of these torques is equal to the moment of inertia of the body times the angular acceleration.
Example:
- Imagine a spinning top. When you apply a force to its top, it starts spinning faster due to the torque generated. The rate at which the spinning speed increases is determined by the moment of inertia of the top and the force applied.
- Angular Acceleration (α):
- The term angular acceleration (α) refers to how quickly the angular velocity of an object changes over time.
- It is a vector quantity that can be positive or negative, indicating the direction of rotation (clockwise or counterclockwise).
- For example, if a gyroscope is spinning counterclockwise and experiences a torque in the clockwise direction, its angular acceleration would be negative.
- Newton's Second Law for Rotation:
- Newton's second law for rotation deals with how torque, moment of inertia, and rotational kinematics are related.
- This law is crucial in solving problems involving the rotation of rigid bodies when forces are applied.
- It essentially explains the relationship between force, moment of inertia, and angular acceleration.
- Rotational Dynamics:
- Rotational dynamics is a branch of physics that focuses on the motion of objects rotating around an axis.
- It helps us understand how forces and torques affect the rotational motion of objects.
- By applying the principles of rotational dynamics, we can analyze and solve various rotational motion problems.
- Deriving Newton's Second Law for Rotation in Vector Form:
- Similar to Newton's second law for translational motion (F=ma), there is an equivalent law for rotational motion relating torque to angular acceleration.
- This equation helps us understand how rotational forces influence the angular acceleration of an object.
- By using vectors to represent torque and angular acceleration, we can analyze rotational motion more effectively.
Understanding Newton's Second Law in Rotational Motion
Key Concept: Newton's Second Law for Rotational Motion
Understanding the Equations
Implications of the Equation
Visualizing the Concept
Understanding Rotational Dynamics Equations
- Firstly, before delving into applying rotational dynamics equations in practical scenarios, it's essential to grasp a problem-solving approach tailored for such situations.
- When faced with rotational dynamics problems, it's crucial to:
- Assess the scenario to identify the involvement of torque and mass in the rotation. Create a detailed sketch of the situation for clarity.
- Pinpoint the system under consideration.
- Create a comprehensive free-body diagram illustrating and labeling all external forces acting on the system of interest.
- Locate the pivot point. In cases of equilibrium, choose the pivot point that simplifies the problem-solving process the most.
- Utilize the equation Σiτi = Iα (the rotational counterpart of Newton’s second law) to tackle the problem. Ensure the accurate application of the moment of inertia and consider the torque concerning the rotation point.
- Always verify the solution's reasonableness post-solving.
Illustrative Example
Let's consider a scenario where a father pushes a playground merry-go-round. He exerts a force of 250 N at the edge of a 50.0-kg merry-go-round with a 1.50-m radius. We aim to calculate the angular acceleration generated in two cases:
- When the merry-go-round is unoccupied.
- When an 18.0-kg child sits 1.25 m away from the center.
This merry-go-round is assumed to be a uniform disk with minimal friction.
- Concept of Torque and Angular Acceleration:
When a father pushes a playground merry-go-round at its edge perpendicular to its radius, he applies a force that results in torque, causing the merry-go-round to rotate. The net torque, represented by the sum of individual torques, is directly proportional to the moment of inertia and angular acceleration.
- Calculating Moment of Inertia:
The moment of inertia of a solid disk, such as the playground merry-go-round, is a measure of its resistance to rotation. It is calculated using the formula I = 1/2 * M * R^2, where M is the mass of the object and R is the radius.
- Determining Net Torque:
When calculating the net torque acting on the merry-go-round, it is important to consider the applied force, the distance from the axis of rotation, and the angle between the force and the radius. In this case, the net torque is found to be 375.0 N-m.
- Angular Acceleration Calculation:
By dividing the net torque by the moment of inertia, the angular acceleration of the merry-go-round is determined to be 6.67 rad/s^2. This value represents how quickly the merry-go-round accelerates in its rotational motion.
Understanding Angular Acceleration and Moment of Inertia
Calculation of Torque and Angular Acceleration
In this scenario, we are examining the concept of torque and angular acceleration in a physical system. Torque (represented by τ) is calculated by multiplying the radius (r) by the force (F) acting perpendicular to the radius. For instance, if r = 1.50m and F = 250.0N, the resulting torque is 375.0N-m.
After substituting the known values into the formula for angular acceleration (α = τ/I), where I represents the moment of inertia, we find the angular acceleration to be 6.67 rad/s2.
Moment of Inertia Calculation
We anticipate a lower angular acceleration due to the increased moment of inertia when a child is added to the merry-go-round system. The total moment of inertia (I) is the sum of the individual moments of inertia of the merry-go-round and the child.
Total Moment of Inertia Calculation
To find the total moment of inertia, we first determine the child's moment of inertia (Ic) by approximating the child as a point mass at a distance from the axis. Substituting the values, we find Ic to be 28.13 kg-m2.
Final Angular Acceleration Calculation
The total moment of inertia (I) is found to be 84.38 kg-m2. Substituting the values into the formula for angular acceleration, we get an angular acceleration of 4.44 rad/s2.
Summary Notes on Rotational Dynamics
Moment of Inertia
- The moment of inertia, denoted by I, is a measure of an object's resistance to changes in its rotation speed.
- For example, when a child is on a merry-go-round, the moment of inertia is 84.38 kg·m2.
Angular Acceleration
- Angular acceleration, represented by α, is the rate of change of angular velocity over time.
- When a father pushes a merry-go-round, the angular acceleration is 4.44 rad/s2 with a torque of 375.0 N·m.
Comparison of Angular Acceleration
- Angular acceleration is lower when the child is on the merry-go-round compared to when it is empty due to higher moment of inertia.
- Friction being neglected results in relatively large angular accelerations.
Practical Example
- If the father pushes for 2.00 seconds, the merry-go-round reaches 13.3 rad/s empty and 8.89 rad/s with the child, equivalent to 2.12 rev/s and 1.41 rev/s respectively.
- The father's speed while pushing would be approximately 50 km/h in the former case.
Check Your Understanding
- A jet engine's fan blades with a moment of inertia of 30.0 kg·m2 accelerate to 20 rev/s in 10 seconds.
- Students are tasked with calculating the torque needed for this acceleration and for bringing the blades to a stop in 20 seconds.
Summary
- Newton's second law for rotation states that the total torque on a rotating system around a fixed axis is equal to the product of the moment of inertia and the angular acceleration, analogous to Newton's second law of linear motion.
- In the vector form of Newton's second law for rotation, the torque vector points in the same direction as the angular acceleration. When the angular acceleration is positive, the torque is also positive, and when the angular acceleration is negative, the torque is negative.
Conceptual Questions
If you were to stop a spinning wheel with a constant force, where on the wheel would you apply the force to produce the maximum negative acceleration?
When a rod is pivoted about one end and forces F and -F are applied to it, the rod will not rotate if the forces are equal in magnitude and opposite in direction.
In Newton's second law for rotation, the concept of torque and angular acceleration is crucial for understanding how rotating systems behave. When a force is applied to a spinning wheel to stop it, the force should be applied at the point farthest from the center of rotation to generate the maximum negative acceleration. For the scenario of a rod pivoted at one end with forces F and -F acting on it, the rod will remain stationary if the forces are of equal magnitude and opposite in direction. This equilibrium of forces prevents the rod from rotating around its pivot point.Understanding these principles helps us grasp the relationship between torque, angular acceleration, and rotational motion in various mechanical systems.
Angular Acceleration and Torque Problems
Concept of Angular Acceleration:
- Angular acceleration refers to the rate of change of angular velocity of an object over time.
- It is calculated in radians per second squared (rad/s2).
- For example, when a flywheel starts from rest and acquires angular velocity due to a torque, it experiences angular acceleration.
Calculation of Torque:
- Torque is the rotational equivalent of force, causing an object to rotate around an axis.
- It is calculated as the product of the force applied and the distance from the axis of rotation.
- For instance, in the case of pressing a steel axe against a rotating grindstone, torque plays a crucial role.
Frictional Forces and Angular Acceleration:
- Frictional forces can affect the angular acceleration of rotating objects.
- Opposing frictional forces reduce the net torque applied to an object, leading to changes in angular acceleration.
- Calculating angular acceleration in the presence of friction is essential for understanding the dynamics of rotating systems.
Application of Concepts:
- Understanding angular acceleration and torque is crucial in various real-world scenarios such as machinery operation and vehicle dynamics.
- Practical examples like the grindstone and flywheel problems help in applying theoretical concepts to solve engineering challenges.
Angular Momentum and Torque
- Moment of Inertia and Torque Relationship:
- Moment of inertia (I) is a measure of an object's resistance to changes in its rotational motion.
- When a torque is applied to a rigid body, it causes angular acceleration, which is directly proportional to the torque and inversely proportional to the moment of inertia.
- For example, if a wheel with a moment of inertia of 4.0 kg·m² achieves an angular velocity of 20.0 rad/s in 10.0 s, the applied torque can be calculated using the formula τ = Iα.
- Applied torque in this scenario is 8.0 N·m.
- Angular Velocity and Friction:
- When a torque is applied to a grinding wheel with a moment of inertia of 20.0 kg·m², it accelerates to a certain angular velocity after the torque is removed.
- The angular velocity and angular displacement can be calculated based on the applied torque and time.
- Frictional Torque and Rotational Motion:
- A flywheel with a moment of inertia of 100.0 kg·m² is brought to rest by friction, resulting in a frictional torque of 43.6 N·m.
- Frictional torque is essential in understanding how rotational motion is affected by external forces.
- Rotational Dynamics in Practical Scenarios:
- Practical examples involving cylindrical grinding wheels and electric motors demonstrate the application of torque in real-world situations.
- Calculations involving torque, angular velocity, and friction help in determining the necessary forces and motions in rotational systems.
Physics Concepts Explained
Rotation of Earth
Initial State of Earth
When Earth was first created, it was not rotating.
Angular Acceleration
The angular acceleration of Earth was 1.4×10^-10 rad/s^2 during the 6-day period.
Applied Torque
A torque of 1.36×10^28 N·m was applied to Earth during this time.
Force Tangent to Earth
A force of 2.1×10^21 N tangent to Earth's equator would produce this torque.
Pulley Mechanics
Inclined Plane Scenario
Block and Inclined Plane
A 3 kg block sliding down an inclined plane at an angle of 45 degrees with a pulley system at the top.
Friction and Acceleration
The coefficient of kinetic friction on the plane is 0.4, determining the acceleration of the block.
Physics Concepts Summary
Acceleration of a Cart
- The acceleration of a cart moving across a table top is determined by neglecting friction.
- Given data: mass of cart (m1 = 2.0kg), mass of block (m2 = 4.0kg), moment of inertia (I = 0.4kg-m^2), and radius (r = 20cm).
- To find the acceleration of the cart, consider the forces acting on the system.
Linear Acceleration of a Rod
- A uniform rod of specific mass and length is held vertically by two strings of negligible mass.
- When the string is cut, the free end of the stick experiences linear acceleration.
- Calculate the linear acceleration immediately after the string is cut at the free end and at the middle of the stick.
Speed and Acceleration of a Metal Disk
- A thin stick attached to the rim of a metal disk is free to rotate around a horizontal axis.
- When released, the stick transitions from a horizontal to a vertical position, affecting the speed and acceleration of the disk's center.
- Determine the speed of the center of the disk when the stick is vertical.
- Find the acceleration of the disk's center at the moment the stick is released and as it passes through the vertical position.
Key Concepts
- Length (L): Represents a measurement of 0.5 meters.
- Mass (M): Indicates a value of 2.0 kilograms.
- Distance (R): Refers to a distance of 0.3 meters.
Glossary
- OpenStax University Physics
- Authored by: OpenStax CNX
- Located at: Link
- License: CC BY: Attribution
- License Terms: Download for free at Link